answer：Alright, so I have this problem here: "Given a linear function y = kx + 3 (k ≠ 0), where the function values of y increase as the values of x increase. When x = -2, the function value y can be ________." Hmm, okay, let's break this down step by step.First, I know that a linear function is generally written as y = mx + b, where m is the slope and b is the y-intercept. In this case, the function is given as y = kx + 3, so k is the slope, and 3 is the y-intercept.Now, the problem mentions that the function values of y increase as the values of x increase. That means that as x gets larger, y also gets larger. In terms of the slope, this tells me that the slope must be positive because a positive slope means the function is increasing. If the slope were negative, y would decrease as x increases, which contradicts the given information.So, I can conclude that k > 0. That's an important piece of information because it restricts the possible values of k.Next, the question asks for the function value y when x = -2. To find this, I need to substitute x = -2 into the equation y = kx + 3. Let's do that:y = k*(-2) + 3Simplifying that, we get:y = -2k + 3Now, since k is positive, the term -2k will be negative. So, y = 3 - 2k. This means that y will be less than 3 because we're subtracting a positive number from 3.But the problem doesn't specify a particular value for k, just that k is not zero and positive. So, y can take on any value less than 3, depending on the value of k. For example, if k = 1, then y = 3 - 2*1 = 1. If k = 2, then y = 3 - 2*2 = -1. If k = 0.5, then y = 3 - 2*0.5 = 2.Wait a minute, but k can't be zero, so k has to be greater than zero, but it can be any positive number, right? So, y can be any real number less than 3. That means y can be 1, -1, 2, or any other number less than 3, depending on the value of k.But the question says, "the function value y can be ________." It doesn't specify to find all possible values or to express it in terms of k. It just asks for a possible value. So, I think I can choose any value less than 3 as a possible answer.Let me test this with a couple of examples to make sure I'm on the right track.If k = 1, then y = 3 - 2*1 = 1.If k = 2, then y = 3 - 2*2 = -1.If k = 0.5, then y = 3 - 2*0.5 = 2.If k = 3, then y = 3 - 2*3 = -3.Okay, so y can be 1, -1, 2, -3, etc., as long as it's less than 3. That makes sense because k is positive, so the more k increases, the more negative y becomes when x = -2.But wait, what if k is a very small positive number, like 0.1? Then y = 3 - 2*0.1 = 3 - 0.2 = 2.8. So, y is still less than 3, but very close to it.Conversely, if k is a very large positive number, like 100, then y = 3 - 2*100 = 3 - 200 = -197. That's a very large negative number.So, y can range from just below 3 to negative infinity, depending on the value of k. But the question doesn't ask for the range; it just asks for a possible value when x = -2.Therefore, any value less than 3 is acceptable. For simplicity, I could choose y = 1, as in the first example, or y = 2, or y = -1, etc.But to make sure, let me think about this again. The function is y = kx + 3, with k > 0. When x = -2, y = -2k + 3. Since k > 0, -2k is negative, so y = 3 - 2k < 3.Yes, that seems correct. So, y must be less than 3. Therefore, any number less than 3 is a possible value for y when x = -2.But the question says, "the function value y can be ________." It doesn't specify to write an expression or to give a range. It just asks to fill in the blank with any correct answer. So, I think it's acceptable to provide a specific value, like 1, as an example.Alternatively, if I wanted to express it in terms of k, I could write y = 3 - 2k, but the question seems to be asking for a numerical value, not an expression.So, to sum up, since k is positive, y when x = -2 will be less than 3. Therefore, any number less than 3 is a possible value for y. For simplicity, I could choose y = 1 as a possible answer.

answer：First, I need to evaluate the logarithmic expressions separately.For the first term, (lg 0.01), since the base is 10, I recognize that (0.01) is (10^{-2}). Therefore, (lg 0.01 = -2).Next, for the second term, (log_{2}16), I note that (16) is (2^{4}). Thus, (log_{2}16 = 4).Finally, I add the two results together: (-2 + 4 = 2).

answer：Okay, so I have this problem here about finding the number of 12-tuples of real numbers that satisfy a certain equation. The equation is a sum of squares, starting from (1 - x₁)², then (x₁ - x₂)², all the way down to x₁₂². And this whole sum equals 1/13. Hmm, interesting.First, I need to understand what this equation represents. It's a sum of squared differences, kind of like a chain from 1 to x₁ to x₂ and so on until x₁₂, and then x₁₂ squared. So it's like each term is the square of the difference between consecutive variables, except the first term is the difference between 1 and x₁, and the last term is just x₁₂ squared.I remember that when dealing with sums of squares, especially in optimization or equations, the Cauchy-Schwarz inequality can sometimes be useful. Maybe I can apply that here. Let me recall the Cauchy-Schwarz inequality: for any real vectors a and b, the square of their dot product is less than or equal to the product of their magnitudes squared. In mathematical terms, (a·b)² ≤ (a·a)(b·b). Equality holds when a and b are scalar multiples of each other.In this problem, I have a sum of squares on the left-hand side. Maybe I can set up vectors a and b such that their dot product relates to the sum I have. Let me think about how to structure this.Looking at the terms, there are 13 terms in total: (1 - x₁)², (x₁ - x₂)², ..., (x₁₁ - x₁₂)², and x₁₂². So that's 12 differences between the x's plus the first term and the last term, making 13 terms. So, each term is squared, and I can think of each as a component of a vector.If I let vector a be [1, 1, 1, ..., 1] with 13 ones, and vector b be [1 - x₁, x₁ - x₂, x₂ - x₃, ..., x₁₁ - x₁₂, x₁₂], then the sum I have is the dot product of b with itself, which is b·b. So, the equation is b·b = 1/13.Now, applying Cauchy-Schwarz, I have (a·b)² ≤ (a·a)(b·b). Let's compute a·a: since a is a vector of 13 ones, a·a is 13. And b·b is given as 1/13. So, (a·b)² ≤ 13 * (1/13) = 1. Therefore, (a·b)² ≤ 1, which means |a·b| ≤ 1.But what is a·b? Let's compute that. a·b is the sum of the components of a multiplied by the components of b. Since a is all ones, it's just the sum of the components of b. So, a·b = (1 - x₁) + (x₁ - x₂) + (x₂ - x₃) + ... + (x₁₁ - x₁₂) + x₁₂.Let me compute this sum step by step. The first term is (1 - x₁). The next term is (x₁ - x₂), which cancels out the x₁. Then (x₂ - x₃) cancels out x₂, and so on, until the last term is x₁₂. So, when we add all these up, most terms cancel out, and we're left with 1 - x₁ + x₁ - x₂ + x₂ - x₃ + ... + x₁₁ - x₁₂ + x₁₂. All the x terms cancel except the first 1 and the last x₁₂, which also cancels because it's subtracted and then added. Wait, no, actually, let me check that again.Wait, the first term is (1 - x₁), then (x₁ - x₂), so 1 - x₁ + x₁ - x₂ = 1 - x₂. Then adding (x₂ - x₃) gives 1 - x₃, and so on. Each time, the x term cancels with the next term. So, after adding all 13 terms, we end up with 1 - x₁₂. Wait, no, because the last term is x₁₂, not (x₁₁ - x₁₂). Wait, no, hold on.Wait, the last term is x₁₂, which is separate from the differences. So, actually, the sum is (1 - x₁) + (x₁ - x₂) + ... + (x₁₁ - x₁₂) + x₁₂. Let's compute this:Start with 1 - x₁.Add x₁ - x₂: 1 - x₁ + x₁ - x₂ = 1 - x₂.Add x₂ - x₃: 1 - x₂ + x₂ - x₃ = 1 - x₃.Continue this way until you add x₁₀ - x₁₁: 1 - x₁₀ + x₁₀ - x₁₁ = 1 - x₁₁.Then add x₁₁ - x₁₂: 1 - x₁₁ + x₁₁ - x₁₂ = 1 - x₁₂.Finally, add x₁₂: 1 - x₁₂ + x₁₂ = 1.So, a·b = 1. Therefore, from Cauchy-Schwarz, we have (a·b)² = 1² = 1 ≤ 13 * (1/13) = 1. So, equality holds in Cauchy-Schwarz.I remember that equality in Cauchy-Schwarz occurs if and only if the vectors a and b are linearly dependent, meaning one is a scalar multiple of the other. So, vector b must be a scalar multiple of vector a. Since a is all ones, that means each component of b must be equal. So, 1 - x₁ = x₁ - x₂ = x₂ - x₃ = ... = x₁₁ - x₁₂ = x₁₂.Let me denote this common difference as d. So, 1 - x₁ = d, x₁ - x₂ = d, x₂ - x₃ = d, ..., x₁₁ - x₁₂ = d, and x₁₂ = d.So, we have a sequence where each term is the previous term minus d. Let me write down the relations:1 - x₁ = d ⇒ x₁ = 1 - d.x₁ - x₂ = d ⇒ x₂ = x₁ - d = (1 - d) - d = 1 - 2d.x₂ - x₃ = d ⇒ x₃ = x₂ - d = (1 - 2d) - d = 1 - 3d.Continuing this pattern, we can see that xₖ = 1 - k*d for k = 1, 2, ..., 12.Wait, let's check for k=12: x₁₂ = 1 - 12d.But from the last term, we have x₁₂ = d. So, 1 - 12d = d ⇒ 1 = 13d ⇒ d = 1/13.So, d is 1/13. Therefore, each xₖ = 1 - k*(1/13) = (13 - k)/13.So, x₁ = (13 - 1)/13 = 12/13.x₂ = (13 - 2)/13 = 11/13.Continuing this way, x₁₂ = (13 - 12)/13 = 1/13.Wait, but earlier, we had x₁₂ = d = 1/13. That matches.So, the 12-tuple is uniquely determined as (12/13, 11/13, 10/13, ..., 1/13, 0). Wait, hold on, x₁₂ is 1/13, but according to xₖ = (13 - k)/13, when k=12, x₁₂ = (13 - 12)/13 = 1/13. But in the last term of the sum, we have x₁₂², which is (1/13)². But in the initial setup, x₁₂ = d = 1/13, so that's consistent.Wait, but in the sequence, x₁₂ is 1/13, but in the 12-tuple, the last term is x₁₂, which is 1/13, not 0. Wait, maybe I made a mistake in the counting.Wait, let me check x₁₂ again. From xₖ = 1 - k*d, with d=1/13, so x₁₂ = 1 - 12*(1/13) = (13 - 12)/13 = 1/13. So, x₁₂ is 1/13, not 0. So, the 12-tuple is (12/13, 11/13, 10/13, ..., 2/13, 1/13). So, all terms are positive, decreasing by 1/13 each time.Wait, but in the initial problem, the last term is x₁₂². So, x₁₂ is 1/13, so x₁₂² is (1/13)², which is 1/169. But in the sum, we have 13 terms each equal to d², since each term is d², because all the differences are equal to d. So, 13*d² = 1/13 ⇒ d² = 1/(13*13) ⇒ d = ±1/13.But in our case, we have d = 1 - x₁ = x₁ - x₂ = ... = x₁₂. Since all the differences are equal, and starting from 1, each subsequent x is less than the previous one by d. So, d must be positive, otherwise, the sequence would increase, which would make x₁₂ larger than 1, but x₁₂ is 1/13, which is less than 1. So, d must be positive, so d = 1/13.Therefore, the 12-tuple is uniquely determined as (12/13, 11/13, ..., 1/13). So, there's only one such 12-tuple.Wait, but let me double-check. If I set d = 1/13, then each term in the sum is (1/13)², and there are 13 terms, so 13*(1/13)² = 1/13, which matches the given equation. So, that works.Is there any other possibility? Well, in the Cauchy-Schwarz inequality, equality holds only when the vectors are proportional. Since we've found that proportionality leads to a unique solution, there are no other solutions. Therefore, there's exactly one such 12-tuple.So, the number of 12-tuples is 1.**Final Answer**The number of such 12-tuples is boxed{1}.

answer：Okay, so I need to prove that for any positive integer ( n ), the number ( n^2 + 8n + 15 ) is not divisible by ( n + 4 ). Hmm, let's see. I remember that when dealing with divisibility, especially with polynomials, polynomial division or factoring might be helpful. Maybe I can try dividing ( n^2 + 8n + 15 ) by ( n + 4 ) and see what happens.Let me set up the division. If I divide ( n^2 + 8n + 15 ) by ( n + 4 ), what would the quotient and remainder be? I think polynomial long division is the way to go here.So, dividing ( n^2 ) by ( n ) gives me ( n ). Then, I multiply ( n + 4 ) by ( n ) to get ( n^2 + 4n ). Subtracting this from the original polynomial:( (n^2 + 8n + 15) - (n^2 + 4n) = 4n + 15 ).Now, I take ( 4n ) and divide it by ( n ), which gives me 4. Multiply ( n + 4 ) by 4 to get ( 4n + 16 ). Subtracting this from the previous remainder:( (4n + 15) - (4n + 16) = -1 ).So, the division gives me a quotient of ( n + 4 ) and a remainder of ( -1 ). That means:( n^2 + 8n + 15 = (n + 4)(n + 4) - 1 ).Wait, that simplifies to:( n^2 + 8n + 15 = (n + 4)^2 - 1 ).Hmm, interesting. So, ( n^2 + 8n + 15 ) is equal to ( (n + 4)^2 - 1 ). Now, if I want to check if ( n + 4 ) divides ( n^2 + 8n + 15 ), I can substitute ( k = n + 4 ). Then, the expression becomes ( k^2 - 1 ).So, the question is whether ( k ) divides ( k^2 - 1 ). Let's see:( frac{k^2 - 1}{k} = k - frac{1}{k} ).Since ( k ) is an integer (because ( n ) is a positive integer, so ( k = n + 4 ) is also an integer greater than 4), ( frac{1}{k} ) is not an integer. Therefore, ( k - frac{1}{k} ) is not an integer, which means ( k ) does not divide ( k^2 - 1 ).Wait, but ( k^2 - 1 ) can be factored as ( (k - 1)(k + 1) ). So, ( frac{(k - 1)(k + 1)}{k} ). For this to be an integer, ( k ) must divide the product ( (k - 1)(k + 1) ). However, since ( k ) and ( k - 1 ) are consecutive integers, they are coprime. Similarly, ( k ) and ( k + 1 ) are consecutive and coprime. Therefore, ( k ) cannot divide either ( k - 1 ) or ( k + 1 ), so it cannot divide their product.Therefore, ( k ) does not divide ( k^2 - 1 ), which means ( n + 4 ) does not divide ( n^2 + 8n + 15 ) for any positive integer ( n ).I think that makes sense. Let me just double-check with a specific example. Let's take ( n = 1 ):( n^2 + 8n + 15 = 1 + 8 + 15 = 24 ).( n + 4 = 5 ).Does 5 divide 24? No, because 24 divided by 5 is 4.8, which is not an integer.Another example, ( n = 2 ):( n^2 + 8n + 15 = 4 + 16 + 15 = 35 ).( n + 4 = 6 ).Does 6 divide 35? No, because 35 divided by 6 is approximately 5.833, which is not an integer.One more example, ( n = 3 ):( n^2 + 8n + 15 = 9 + 24 + 15 = 48 ).( n + 4 = 7 ).Does 7 divide 48? No, because 48 divided by 7 is approximately 6.857, which is not an integer.Okay, these examples support the conclusion. So, it seems that ( n + 4 ) does not divide ( n^2 + 8n + 15 ) for positive integers ( n ).I think I've got it. The key was to perform polynomial division and realize that the remainder is -1, which means it's not divisible. Alternatively, factoring the expression and analyzing the divisibility using properties of consecutive integers also leads to the same conclusion.

answer：Alright, so I've got this problem here: find all triplets of natural numbers (x, y, z) such that x² + y² = 3·2016ᶻ + 77. Hmm, okay, let's break this down step by step.First off, natural numbers mean x, y, z are positive integers, right? So, z has to be at least 1, but wait, actually, natural numbers sometimes include zero depending on the context. I should check if z can be zero. If z is zero, then 2016ᶻ is 1, so the equation becomes x² + y² = 3·1 + 77 = 80. That seems manageable. But let's see if z can be zero or if it's strictly positive.But before that, let's look at the equation: x² + y² = 3·2016ᶻ + 77. 2016 is a big number, and it's being raised to the power z. So, depending on z, the right-hand side can get really large. Since x and y are natural numbers, their squares have to add up to this value. So, z can't be too large because otherwise, 3·2016ᶻ + 77 would be way too big for x and y to be natural numbers.Let me think about the possible values of z. Maybe starting with small z and seeing if the equation holds.First, let's try z = 0. Then, as I said, the equation becomes x² + y² = 80. Let's see what pairs (x, y) satisfy this. So, x and y are natural numbers, so let's list the squares less than or equal to 80:1² = 12² = 43² = 94² = 165² = 256² = 367² = 498² = 649² = 81 (which is already more than 80)So, possible squares are 1, 4, 9, 16, 25, 36, 49, 64.Now, let's see which pairs add up to 80.Start with 64: 80 - 64 = 16, which is 4². So, (8, 4) and (4, 8).Next, 49: 80 - 49 = 31, which isn't a perfect square.36: 80 - 36 = 44, not a square.25: 80 - 25 = 55, not a square.16: 80 - 16 = 64, which is 8². So, (4, 8) and (8, 4), which we already have.9: 80 - 9 = 71, not a square.4: 80 - 4 = 76, not a square.1: 80 - 1 = 79, not a square.So, for z = 0, the solutions are (8, 4, 0) and (4, 8, 0).Okay, that's z = 0. Now, let's try z = 1. Then, the equation becomes x² + y² = 3·2016 + 77. Let's compute that:3·2016 = 60486048 + 77 = 6125So, x² + y² = 6125. Hmm, that's a much bigger number. Let's see if we can find natural numbers x and y such that their squares add up to 6125.First, let's find the approximate square root of 6125 to get an idea of how big x and y can be.√6125 ≈ 78.25So, x and y can be up to 78.But 6125 is a large number, so maybe there's a pattern or a way to factor it to find possible x and y.Wait, 6125. Let's factor it:6125 ÷ 25 = 245245 ÷ 5 = 4949 ÷ 7 = 7So, 6125 = 5³ × 7²Hmm, interesting. So, 6125 is 5³ × 7².Now, x² + y² = 5³ × 7²I wonder if there's a way to express this as a sum of squares. Maybe using the identity that if a number can be expressed as a sum of squares, then its prime factors of the form 4k+3 must have even exponents. Let's check:5 is 1 mod 4, so it's okay. 7 is 3 mod 4, and its exponent is 2, which is even. So, yes, 6125 can be expressed as a sum of two squares.Now, to find all pairs (x, y) such that x² + y² = 6125.One approach is to look for squares that are less than or equal to 6125 and see if their complement is also a square.But that might take a while. Maybe there's a smarter way.Alternatively, since 6125 = 25 × 245, and 245 = 49 × 5, so 6125 = 25 × 49 × 5 = 5³ × 7².Wait, maybe we can use the fact that 5 and 7 can be expressed as sums of squares.5 = 1² + 2²7 can't be expressed as a sum of two squares, but 7² = 49 = 0² + 7², but 0 isn't a natural number. Hmm.Alternatively, maybe we can use the identity for multiplying sums of squares.If we have two numbers expressed as sums of squares, their product can also be expressed as a sum of squares.But I'm not sure if that helps here.Wait, let's try to find x and y such that x² + y² = 6125.Let me think about possible squares near 6125.78² = 60846125 - 6084 = 41, which isn't a perfect square.77² = 59296125 - 5929 = 196, which is 14².So, 77² + 14² = 5929 + 196 = 6125.Great, so (77, 14) is a solution.Similarly, (14, 77) is also a solution.Now, let's see if there are other pairs.Let's try 70² = 49006125 - 4900 = 1225, which is 35².So, 70² + 35² = 4900 + 1225 = 6125.So, (70, 35) and (35, 70) are solutions.Are there any more?Let's check 65² = 42256125 - 4225 = 1900, which isn't a perfect square.60² = 36006125 - 3600 = 2525, not a square.55² = 30256125 - 3025 = 3100, not a square.50² = 25006125 - 2500 = 3625, not a square.45² = 20256125 - 2025 = 4100, not a square.40² = 16006125 - 1600 = 4525, not a square.35² = 1225We already have that with 70².30² = 9006125 - 900 = 5225, not a square.25² = 6256125 - 625 = 5500, not a square.20² = 4006125 - 400 = 5725, not a square.15² = 2256125 - 225 = 5900, not a square.10² = 1006125 - 100 = 6025, not a square.5² = 256125 - 25 = 6100, not a square.So, it seems the only solutions for z = 1 are (77, 14, 1), (14, 77, 1), (70, 35, 1), and (35, 70, 1).Now, let's check z = 2.So, z = 2: x² + y² = 3·2016² + 77.First, compute 2016²:2016 × 2016. Let's compute that.2016 × 2000 = 4,032,0002016 × 16 = 32,256So, total is 4,032,000 + 32,256 = 4,064,256Then, 3·4,064,256 = 12,192,768Add 77: 12,192,768 + 77 = 12,192,845So, x² + y² = 12,192,845Now, let's see if this is possible.First, find the approximate square root: √12,192,845 ≈ 3491So, x and y can be up to 3491.But 12,192,845 is a huge number. Let's see if it's possible to express it as a sum of two squares.Again, let's factor it.12,192,845First, check if it's divisible by 5: last digit is 5, so yes.12,192,845 ÷ 5 = 2,438,569Now, check 2,438,569.Let's see if it's divisible by 7: 2,438,569 ÷ 7 ≈ 348,367, which is not an integer.Wait, 7 × 348,367 = 2,438,569? Let's check:7 × 348,367 = 7 × 300,000 = 2,100,0007 × 48,367 = 338,569So, total is 2,100,000 + 338,569 = 2,438,569. Yes, so 2,438,569 = 7 × 348,367Now, check 348,367.Is this divisible by 7? 348,367 ÷ 7 ≈ 49,766.714, not an integer.Check divisibility by 13: 348,367 ÷ 13 ≈ 26,797.461, not integer.Check 3: 3 + 4 + 8 + 3 + 6 + 7 = 31, not divisible by 3.Check 5: doesn't end with 0 or 5.Check 11: 3 - 4 + 8 - 3 + 6 - 7 = 3 -4= -1, -1 +8=7, 7-3=4, 4+6=10, 10-7=3, not divisible by 11.Hmm, maybe it's prime? Not sure.But regardless, 12,192,845 = 5 × 7 × 348,367Now, to express this as a sum of two squares, we need to check the prime factors.Primes of the form 4k+3 must have even exponents.Here, 5 is 1 mod 4, 7 is 3 mod 4, and 348,367 is unknown.If 348,367 is a prime, and if it's 3 mod 4, then its exponent is 1, which is odd, making it impossible to express as a sum of two squares.Alternatively, if 348,367 is composite and its prime factors of the form 4k+3 have even exponents, then it might be possible.But without knowing the factorization, it's hard to say.Alternatively, maybe we can use the fact that 2016 is divisible by 7, so 2016^z is divisible by 7^z.So, 3·2016^z + 77 is divisible by 7, since 2016 is divisible by 7, and 77 is divisible by 7.So, x² + y² must be divisible by 7.Now, quadratic residues modulo 7 are 0, 1, 2, and 4.So, possible squares modulo 7 are 0, 1, 2, 4.So, x² + y² mod 7 can be:0 + 0 = 00 + 1 = 10 + 2 = 20 + 4 = 41 + 1 = 21 + 2 = 31 + 4 = 52 + 2 = 42 + 4 = 64 + 4 = 1So, the possible sums modulo 7 are 0,1,2,3,4,5,6.But x² + y² must be congruent to 0 mod 7.Looking at the possible sums, only 0 + 0 gives 0 mod 7.So, both x and y must be divisible by 7.So, x = 7a, y = 7b, where a and b are natural numbers.Substituting back into the equation:(7a)² + (7b)² = 3·2016^z + 7749a² + 49b² = 3·2016^z + 7749(a² + b²) = 3·2016^z + 77Now, let's divide both sides by 7:7(a² + b²) = (3·2016^z)/7 + 11But 2016 is divisible by 7, so 2016^z is divisible by 7^z.So, (3·2016^z)/7 = 3·2016^(z-1)·288, since 2016/7 = 288.Wait, 2016 = 7 × 288, so 2016^z = 7^z × 288^z.So, (3·2016^z)/7 = 3·7^(z-1)·288^zSo, 7(a² + b²) = 3·7^(z-1)·288^z + 11Now, the left side is divisible by 7, and the right side must also be divisible by 7.Looking at the right side: 3·7^(z-1)·288^z is divisible by 7^(z-1). The other term is 11, which is not divisible by 7.So, for the entire right side to be divisible by 7, 11 must be congruent to 0 mod 7, which it's not. Therefore, there's a contradiction unless z-1 = 0, meaning z = 1.Wait, let's see:If z = 1, then 3·7^(1-1)·288^1 + 11 = 3·1·288 + 11 = 864 + 11 = 875So, 7(a² + b²) = 875Divide both sides by 7: a² + b² = 125Okay, that's manageable.So, for z = 1, we have a² + b² = 125.Now, let's find all pairs (a, b) such that a² + b² = 125.Again, let's list squares less than or equal to 125:1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121Now, let's see which pairs add up to 125.Start with 121: 125 - 121 = 4, which is 2². So, (11, 2) and (2, 11).Next, 100: 125 - 100 = 25, which is 5². So, (10, 5) and (5, 10).81: 125 - 81 = 44, not a square.64: 125 - 64 = 61, not a square.49: 125 - 49 = 76, not a square.36: 125 - 36 = 89, not a square.25: 125 - 25 = 100, which is 10². Already covered.16: 125 - 16 = 109, not a square.9: 125 - 9 = 116, not a square.4: 125 - 4 = 121, which is 11². Already covered.1: 125 - 1 = 124, not a square.So, the pairs are (11, 2), (2, 11), (10, 5), (5, 10).Therefore, the corresponding (x, y) are:For (11, 2): x = 7×11 = 77, y = 7×2 = 14For (2, 11): x = 14, y = 77For (10, 5): x = 70, y = 35For (5, 10): x = 35, y = 70So, for z = 1, we have four solutions: (77, 14, 1), (14, 77, 1), (70, 35, 1), (35, 70, 1).Now, let's check z = 2 again.Earlier, we saw that 12,192,845 = 5 × 7 × 348,367But since 348,367 is not divisible by 7, and we don't know its factors, it's hard to say if it can be expressed as a sum of two squares. However, from the earlier step, when we tried to express x² + y² = 12,192,845, we saw that both x and y must be divisible by 7, leading to a² + b² = 12,192,845 / 49 ≈ 248,837.653, which is not an integer. Wait, that can't be right.Wait, no, earlier, we had:49(a² + b²) = 3·2016^z + 77For z = 2, 3·2016² + 77 = 12,192,845So, 49(a² + b²) = 12,192,845Divide both sides by 49: a² + b² = 12,192,845 / 49 ≈ 248,837.653But 12,192,845 ÷ 49 = 248,837.653, which is not an integer. Wait, that can't be right because 49 × 248,837 = 12,192,813, and 12,192,845 - 12,192,813 = 32, so it's not divisible by 49.Wait, that's a problem. So, 3·2016² + 77 must be divisible by 49?Wait, let's check:3·2016² + 772016 is divisible by 7, so 2016² is divisible by 49.So, 3·2016² is divisible by 49.77 is divisible by 7, but not by 49.So, 3·2016² + 77 ≡ 0 + 77 mod 4977 mod 49 = 28So, 3·2016² + 77 ≡ 28 mod 49But 49(a² + b²) must be ≡ 0 mod 49, but 3·2016² + 77 ≡ 28 mod 49, which is not 0. Therefore, there's a contradiction.This means that for z = 2, there are no solutions because 3·2016² + 77 is not divisible by 49, which is required for x and y to be multiples of 7.Therefore, z cannot be 2.Similarly, for z > 2, 3·2016^z + 77 will be even larger, and likely not divisible by 49, making it impossible for x and y to be natural numbers.Wait, let's check z = 3 just to be sure.z = 3: x² + y² = 3·2016³ + 772016³ is 2016 × 2016 × 2016, which is a huge number, but let's see modulo 49.2016 ≡ 0 mod 49, since 2016 = 49 × 41.142... Wait, no, 49 × 41 = 2009, so 2016 - 2009 = 7, so 2016 ≡ 7 mod 49.So, 2016 ≡ 7 mod 49Therefore, 2016³ ≡ 7³ mod 497³ = 343343 ÷ 49 = 7, so 343 ≡ 0 mod 49So, 2016³ ≡ 0 mod 49Thus, 3·2016³ ≡ 0 mod 4977 ≡ 77 mod 49 ≡ 28 mod 49So, 3·2016³ + 77 ≡ 0 + 28 ≡ 28 mod 49Again, not divisible by 49, so 49(a² + b²) ≡ 0 mod 49, but the right side is 28 mod 49, which is not 0. Therefore, no solutions for z = 3.Similarly, for z ≥ 2, 3·2016^z + 77 ≡ 28 mod 49, which is not 0, so no solutions.Therefore, the only possible values for z are 0 and 1.We've already found the solutions for z = 0 and z = 1.So, compiling all the solutions:For z = 0:(8, 4, 0) and (4, 8, 0)For z = 1:(77, 14, 1), (14, 77, 1), (70, 35, 1), (35, 70, 1)I think that's all the possible triplets.Just to double-check, let's verify one of the solutions for z = 1.Take (77, 14, 1):77² + 14² = 5929 + 196 = 61253·2016 + 77 = 6048 + 77 = 6125Yes, that works.Another one: (70, 35, 1)70² + 35² = 4900 + 1225 = 6125Same as above.For z = 0: (8, 4, 0)8² + 4² = 64 + 16 = 803·1 + 77 = 3 + 77 = 80Yes, that works too.So, all the solutions check out.

answer：Okay, so I have this function f(x) defined piecewise. It's (1/4)x + 1 when x is less than or equal to 1, and it's ln(x) when x is greater than 1. I need to find the range of real numbers a such that the equation f(x) = a x has exactly two distinct real roots. Hmm, okay.Let me start by understanding what this means. The equation f(x) = a x is asking for the points where the function f(x) intersects the line y = a x. So, I need to find the values of a where this line intersects f(x) exactly twice. Since f(x) is piecewise, I should probably consider each piece separately and then see how they interact.First, let's consider the case when x ≤ 1. Here, f(x) = (1/4)x + 1. So, setting this equal to a x gives (1/4)x + 1 = a x. Let me rearrange this equation to solve for x. Subtract a x from both sides: (1/4 - a)x + 1 = 0. Then, (1/4 - a)x = -1. So, x = -1 / (1/4 - a). Simplify that: x = 1 / (a - 1/4). Wait, so for x ≤ 1, the solution is x = 1 / (a - 1/4). But this solution must satisfy x ≤ 1. So, 1 / (a - 1/4) ≤ 1. Let me solve this inequality. First, note that the denominator a - 1/4 cannot be zero, so a ≠ 1/4. Also, depending on the sign of the denominator, the inequality will behave differently. Let's consider two cases:1. If a - 1/4 > 0, which means a > 1/4, then we can multiply both sides by (a - 1/4) without changing the inequality direction: 1 ≤ a - 1/4. So, 1 + 1/4 ≤ a, which means a ≥ 5/4. But wait, if a > 1/4, then a ≥ 5/4 would be the condition for x ≤ 1. But 5/4 is 1.25, which is greater than 1, but x is supposed to be ≤ 1. So, does that mean that for a ≥ 5/4, the solution x = 1 / (a - 1/4) is less than or equal to 1? Let me plug in a = 5/4. Then x = 1 / (5/4 - 1/4) = 1 / (4/4) = 1. So, when a = 5/4, x = 1. For a > 5/4, x would be less than 1, which is still valid because x ≤ 1. So, for a > 1/4, the solution x = 1 / (a - 1/4) is valid only if a ≥ 5/4? Wait, no, that doesn't make sense. Let me think again.Wait, if a > 1/4, then (a - 1/4) is positive, so x = 1 / (a - 1/4). For x ≤ 1, we have 1 / (a - 1/4) ≤ 1. So, 1 ≤ a - 1/4, which implies a ≥ 5/4. So, for a ≥ 5/4, the solution x = 1 / (a - 1/4) is ≤ 1. But if a is between 1/4 and 5/4, then 1 / (a - 1/4) would be greater than 1, which is not allowed because in this case, x must be ≤ 1. So, for a in (1/4, 5/4), the equation f(x) = a x in the region x ≤ 1 has no solution because x would have to be greater than 1, which contradicts x ≤ 1. Wait, that seems a bit confusing. Let me think of it differently. If a is exactly 1/4, then the equation becomes (1/4)x + 1 = (1/4)x, which simplifies to 1 = 0, which is impossible. So, no solution when a = 1/4. If a < 1/4, then (1/4 - a) is positive, so x = -1 / (1/4 - a). Let's compute that: x = -1 / (positive number) = negative number. So, x is negative. But in the region x ≤ 1, x can be negative, so that's okay. So, for a < 1/4, we have a solution x = 1 / (a - 1/4), which is negative because a - 1/4 is negative. So, for a < 1/4, we have one solution in x ≤ 1.Wait, so summarizing:- For a < 1/4: one solution in x ≤ 1.- For a = 1/4: no solution in x ≤ 1.- For a > 1/4: solution x = 1 / (a - 1/4). But this x must be ≤ 1, so 1 / (a - 1/4) ≤ 1. As we saw earlier, this implies a ≥ 5/4. So, for a ≥ 5/4, we have a solution in x ≤ 1, and for 1/4 < a < 5/4, no solution in x ≤ 1.Wait, that seems inconsistent because when a is greater than 1/4, the solution x is positive, but whether it's less than or equal to 1 depends on a. So, for a = 1/2, which is between 1/4 and 5/4, x = 1 / (1/2 - 1/4) = 1 / (1/4) = 4, which is greater than 1, so it's not in x ≤ 1. So, indeed, for a between 1/4 and 5/4, no solution in x ≤ 1. For a ≥ 5/4, x = 1 / (a - 1/4) ≤ 1. So, for a ≥ 5/4, one solution in x ≤ 1.Okay, so that's the first part.Now, moving on to the second piece of f(x): when x > 1, f(x) = ln(x). So, setting this equal to a x: ln(x) = a x.I need to find the number of solutions to ln(x) = a x for x > 1. Let's analyze this equation.Let me define g(x) = ln(x) - a x. We need to find the number of roots of g(x) = 0 for x > 1.Compute the derivative of g(x): g’(x) = 1/x - a.Set derivative equal to zero to find critical points: 1/x - a = 0 => x = 1/a.So, the function g(x) has a critical point at x = 1/a. Now, depending on the value of a, this critical point may lie in the domain x > 1 or not.Case 1: If 1/a > 1, which implies a < 1. So, for a < 1, the critical point is at x = 1/a > 1.Case 2: If 1/a ≤ 1, which implies a ≥ 1. Then, the critical point is at x ≤ 1, which is outside our domain x > 1. So, for a ≥ 1, g’(x) = 1/x - a < 1/1 - a = 1 - a ≤ 0. So, g(x) is decreasing on x > 1.Let me analyze each case.Case 1: a < 1.In this case, g(x) has a critical point at x = 1/a > 1. Let's check the behavior of g(x):- As x approaches 1 from the right: ln(1) = 0, so g(1+) = 0 - a*1 = -a. Since a is positive (we'll assume a > 0 for now; if a is negative, ln(x) is always positive for x > 1, so ln(x) = a x would have no solutions if a is negative because a x would be negative, but ln(x) is positive. So, we can consider a > 0 only.So, as x approaches 1 from the right, g(x) approaches -a.At x = 1/a, which is the critical point, let's compute g(1/a):g(1/a) = ln(1/a) - a*(1/a) = -ln(a) - 1.Now, the behavior as x approaches infinity: ln(x) grows slower than a x, so g(x) approaches -infinity.So, the function g(x) starts at x=1 with g(1) = -a, goes up to a maximum at x=1/a of g(1/a) = -ln(a) -1, and then decreases to -infinity as x approaches infinity.So, for g(x) to have two roots in x > 1, the maximum at x=1/a must be positive. Because if the maximum is positive, then the function crosses the x-axis twice: once before the maximum and once after. If the maximum is zero, it's tangent to the x-axis, so one root. If the maximum is negative, no roots.So, the condition for two roots is that g(1/a) > 0.So, -ln(a) -1 > 0 => -ln(a) > 1 => ln(a) < -1 => a < e^{-1} = 1/e ≈ 0.3679.But wait, in Case 1, a < 1, so 1/e is less than 1, so this is within Case 1.So, for a < 1/e, g(x) has two roots in x > 1.For a = 1/e, g(1/a) = -ln(1/e) -1 = -(-1) -1 = 1 -1 = 0. So, one root.For 1/e < a < 1, g(1/a) = -ln(a) -1. Since a > 1/e, ln(a) > -1, so -ln(a) < 1, so -ln(a) -1 < 0. So, maximum is below zero, so no roots.For a ≥ 1, as discussed earlier, g(x) is decreasing on x > 1, starting from g(1) = -a < 0 and going to -infinity. So, no roots.Wait, but hold on. For a = 0, ln(x) = 0, which has a solution at x=1, but x >1, so no solution. For a negative, as I thought earlier, no solution because ln(x) is positive and a x is negative. So, only for a >0, we can have solutions.So, summarizing:- For a < 1/e: two roots in x >1.- For a =1/e: one root in x >1.- For a >1/e: no roots in x >1.Wait, but hold on, in Case 1, a <1, but the critical point is at x=1/a. So, for a <1/e, 1/a > e, which is still greater than 1, so it's in the domain x >1. So, the analysis holds.So, now, putting it all together.We have two regions: x ≤1 and x >1.In x ≤1:- For a <1/4: one solution in x ≤1.- For a =1/4: no solution.- For 1/4 < a <5/4: no solution.- For a ≥5/4: one solution in x ≤1.In x >1:- For a <1/e: two solutions.- For a =1/e: one solution.- For a >1/e: no solutions.Now, the total number of solutions is the sum of solutions in x ≤1 and x >1.We need exactly two distinct real roots.So, let's consider different ranges of a:1. a <1/4:- x ≤1: one solution.- x >1: two solutions.Total: three solutions. Not desired.2. a =1/4:- x ≤1: no solution.- x >1: two solutions.Total: two solutions. Wait, but hold on, when a=1/4, in x ≤1, the equation becomes (1/4)x +1 = (1/4)x, which simplifies to 1=0, which is impossible, so no solution. In x >1, for a=1/4, which is less than 1/e (~0.3679), so two solutions. So, total two solutions. Hmm, so a=1/4 is a candidate.But wait, let me check. When a=1/4, in x >1, we have two solutions. So, total two solutions. So, a=1/4 is included.3. 1/4 < a <1/e:- x ≤1: no solution.- x >1: two solutions.Total: two solutions. So, in this interval, we have two solutions.Wait, but hold on, 1/4 is approximately 0.25, and 1/e is approximately 0.3679. So, 1/4 < a <1/e is approximately 0.25 < a <0.3679.In this interval, in x ≤1, no solution because a is between 1/4 and 5/4, but in this case, a is less than 5/4, so no solution in x ≤1. In x >1, two solutions. So, total two solutions.4. a =1/e:- x ≤1: no solution (since a=1/e ≈0.3679 <5/4=1.25, so no solution in x ≤1).- x >1: one solution.Total: one solution. Not desired.5. 1/e < a <5/4:- x ≤1: no solution.- x >1: no solution.Total: zero solutions. Not desired.6. a =5/4:- x ≤1: one solution (x=1).- x >1: no solution (since a=5/4 >1/e, so no solution in x >1).Total: one solution. Not desired.7. a >5/4:- x ≤1: one solution (x=1 / (a -1/4) <1).- x >1: no solution.Total: one solution. Not desired.So, putting it all together, the equation f(x)=a x has exactly two distinct real roots when:- a=1/4: two solutions (both in x >1).- 1/4 < a <1/e: two solutions (both in x >1).- a=1/e: one solution.Wait, but hold on, when a=1/4, the solution in x >1 is two, and in x ≤1, none. So, total two.When a is between 1/4 and1/e, same thing: two solutions in x >1, none in x ≤1.When a=1/e, one solution in x >1, none in x ≤1.When a >1/e, no solutions in x >1, but for a ≥5/4, one solution in x ≤1.So, the range of a where the equation has exactly two distinct real roots is a=1/4 and 1/4 <a <1/e. So, combining these, it's [1/4,1/e).Wait, but when a=1/4, we have two solutions in x >1, right? Because a=1/4 <1/e, so two solutions in x >1. So, yes, a=1/4 is included.Therefore, the range is [1/4,1/e).Wait, but let me double-check when a=1/4.At a=1/4, in x ≤1, the equation is (1/4)x +1 = (1/4)x, which simplifies to 1=0, which is impossible. So, no solution in x ≤1.In x >1, a=1/4 <1/e, so two solutions. So, total two solutions.Yes, so a=1/4 is included.Similarly, when a approaches1/e from below, the two solutions in x >1 approach each other and merge at a=1/e, giving one solution.So, the range is [1/4,1/e).Wait, but hold on, when a=1/4, the line y=a x is tangent to the first piece f(x)=(1/4)x +1 at x=1, because f(1)= (1/4)(1)+1=5/4, and y=a x at x=1 is a*1=1/4. Wait, no, that's not tangent. Wait, f(1)=5/4, and y=a x at x=1 is a. So, if a=5/4, then y=5/4 x would pass through (1,5/4), which is the point (1,f(1)). So, for a=5/4, the line passes through (1,f(1)).But in our case, a=1/4, so y=1/4 x at x=1 is y=1/4, but f(1)=5/4. So, they don't intersect at x=1. So, when a=1/4, the line y=1/4 x intersects f(x) only in x >1, twice.Wait, but when a=1/4, in x ≤1, the equation f(x)=a x has no solution because it leads to 1=0. So, only two solutions in x >1.So, yes, a=1/4 is included.Therefore, the range is [1/4,1/e).I think that's the answer.**Final Answer**The range of the real number ( a ) is (boxed{left[ frac{1}{4}, frac{1}{e} right)}).