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上次那个x/(y+z) +y/(z+x) +z/(x+y) =4是难度很大，换一个试试


求解出所有的p^x - y^p = 1, x, y, p >0整数，p 为质数

请找出所有解

3^2 - 2^3 = 1。还有没有别的，我就不知道了，大家接着找。

YWY 写了： 2023年 1月 25日 17:19 3^2 - 2^3 = 1。还有没有别的，我就不知道了，大家接着找。

2^1-1^2=1

这个最适合python。我试了试，x从1到400，y从1到400，p为100之内的质数，只找到上面两组解。

YWY 写了： 2023年 1月 25日 17:19 3^2 - 2^3 = 1。还有没有别的，我就不知道了，大家接着找。

这个题目比我之前想的难，要用到catalan猜想的结果

https://en.wikipedia.org/wiki/Catalan%27s_conjecture

可以把这个结果当做已知结论

（ヅ） 写了： 2023年 1月 25日 16:45 求解出所有的p^x - y^p = 1, x, y, p >0整数，p 为质数

请找出所有解

verdelite 写了： 2023年 1月 25日 20:052^1-1^2=1

（ヅ） 写了： 2023年 1月 25日 22:40 这个题目比我之前想的难，要用到catalan猜想的结果

https://en.wikipedia.org/wiki/Catalan%27s_conjecture

可以把这个结果当做已知结论

要是承认catalan猜想的话，那一楼问题里，不被catalan猜想包含的情况只剩下x=1，所以我们只需额外考虑“p - y^p = 1, y, p >0整数，p 为质数”的情况。当y=1时，得到2^1 - 1^2 = 1的情况。当y > 1时，p = y^p + 1不可能成立，因为2^p > p。

我觉得不需要承认catalan猜想也能证出来，但可能会需要很多推导。

基本题目里面有素数的题目就不好搞，我看着就走

（ヅ） 写了： 2023年 1月 25日 16:45 上次那个x/(y+z) +y/(z+x) +z/(x+y) =4是难度很大，换一个试试

求解出所有的p^x - y^p = 1, x, y, p >0整数，p 为质数

请找出所有解

When p = 2 and x = 1, y = 1 is the only solution. When p = 2 and x > 1, there is no way for 2^x = y^2 = 1 to hold because in this case 2^x - 1 = 3 mod 4 but y^2 is never 3 mod 4.

It remains to consider the case where p is an odd prime (p > 2). Then p^x - y^p = 1 --> y^p = -1 mod p --> y = -1 mod p --> y = kp - 1 where k > 0. Rewrite the equation as (kp - 1)^p + 1 = p^x. Since k divides (kp - 1)^p + 1, we see k divides p^x, so k is a power of p (including p^0). So y = kp - 1 = p^n - 1 with n > 0. So the equation becomes p^x = (p^n - 1)^p + 1. By binomial expansion, we get
p^x = (p^n)^p - p*(p^n)^{p-1} + ... + p*(p^n) - 1 + 1, which simplifies to
p^x = (p^n)^p - p*(p^n)^{p-1} + ... + p*(p^n), with all the binomial coefficients divisible by p (enough to know they are all integers). So we get
p^x = p*p^n*t where t = [(p^n)^p - p*(p^n)^{p-1} + ... + p*(p^n)] /(p*p^n). If we compute t as a quotient term by term (via formula (a + b)/c = a/c + b/c), we see that t is not divisible by p (in fact t = 1 mod p). But t divides p^x. So t must be 1. Hence p^x = p* p^n. Now the equation becomes
p*p^n = (p^n - 1)^p + 1. Let z := p^n, which is at least p, which is at least 3. Using calculus, we see that the function
f(z) := (z - 1)^p + 1 - z*p is strictly increasing in the interval from z = 3 to infinity. When z = 3, the function
g(p) := (3 - 1)^p + 1 - 3p is strictly increasing in the from p = 3 to infinity, with g(3) = 0. Thus the equation
p*p^n = (p^n - 1)^p + 1, with p at least 3 and n at least 1,
holds only when p = 3. This corresponds to the case 3^2 = 2^3 + 1, which is only solution when p is an odd prime.

YWY 写了： 2023年 1月 26日 00:36 When p = 2 and x = 1, y = 1 is the only solution. When p = 2 and x > 1, there is no way for 2^x = y^2 = 1 to hold because in this case 2^x - 1 = 3 mod 4 but y^2 is never 3 mod 4.

It remains to consider the case where p is an odd prime (p > 2). Then p^x - y^p = 1 --> y^p = -1 mod p --> y = -1 mod p --> y = kp - 1 where k > 0. Rewrite the equation as (kp - 1)^p + 1 = p^x. Since k divides (kp - 1)^p + 1, we see k divides p^x, so k is a power of p (including p^0). So y = kp - 1 = p^n - 1 with n > 0. So the equation becomes p^x = (p^n - 1)^p + 1. By binomial expansion, we get
p^x = (p^n)^p - p*(p^n)^{p-1} + ... + p*(p^n) - 1 + 1, which simplifies to
p^x = (p^n)^p - p*(p^n)^{p-1} + ... + p*(p^n), with all the binomial coefficients divisible by p (enough to know they are all integers). So we get
p^x = p*p^n*t where t = [(p^n)^p - p*(p^n)^{p-1} + ... + p*(p^n)] /(p*p^n). If we compute t as a quotient term by term (via formula (a + b)/c = a/c + b/c), we see that t is not divisible by p (in fact t = 1 mod p). But t divides p^x. So t must be 1. Hence p^x = p* p^n. Now the equation becomes
p*p^n = (p^n - 1)^p + 1. Let z := p^n, which is at least p, which is at least 3. Using calculus, we see that the function
f(z) := (z - 1)^p + 1 - z*p is strictly increasing in the interval from z = 3 to infinity. When z = 3, the function
g(p) := (3 - 1)^p + 1 - 3p is strictly increasing in the from p = 3 to infinity, with g(3) = 0. Thus the equation
p*p^n = (p^n - 1)^p + 1, with p at least 3 and n at least 1,
holds only when p = 3. This corresponds to the case 3^2 = 2^3 + 1, which is only solution when p is an odd prime.

对于" t must be 1"有疑问

我自己算过程差不多，p^x不是随机项，而是特定的值，所以不能轻易的消掉，我卡在这一步了.

（ヅ） 写了： 2023年 1月 26日 11:37 对于" t must be 1"有疑问

我自己算过程差不多，p^x不是随机项，而是特定的值，所以不能轻易的消掉，我卡在这一步了.

Note that t = [(p^n)^p - c_1p*(p^n)^{p-1} + ... - c_{p-2}(p^n)^2+ p*(p^n)] /(p*p^n) where c_i is the "p choose i". Observe that p divides c_i for all i = 1, 2, ..., p-1 and c_{p-1} = p. Thus, as p divides c_{p-2}, n > 0 and p > 2, we see that p^2*p^n divides [(p^n)^p - c_1p*(p^n)^{p-1} + ... - c_{p-2}(p^n)^2]. This shows that t is a positive integer and t = 1 mod p. Since t divides p^x, we conclude that t = 1.

YWY 写了： 2023年 1月 25日 23:12 要是承认catalan猜想的话，那一楼问题里，不被catalan猜想包含的情况只剩下x=1，所以我们只需额外考虑“p - y^p = 1, y, p >0整数，p 为质数”的情况。当y=1时，得到2^1 - 1^2 = 1的情况。当y > 1时，p = y^p + 1不可能成立，因为2^p > p。

我觉得不需要承认catalan猜想也能证出来，但可能会需要很多推导。

什么意思？catalan猜想已经被证明了。

FoxMe 写了： 2023年 1月 27日 17:02 什么意思？catalan猜想已经被证明了。

没有吧。我们在楼上说的是，承认catalan猜想（的正确性）的话，一楼的问题可以很容易解决。

Catalan's conjecture (or Mihăilescu's theorem) is a theorem in number theory that was conjectured by the mathematician Eugène Charles Catalan in 1844 and proven in 2002 by Preda Mihăilescu at Paderborn University.

FoxMe 写了： 2023年 1月 27日 17:13 Catalan's conjecture (or Mihăilescu's theorem) is a theorem in number theory that was conjectured by the mathematician Eugène Charles Catalan in 1844 and proven in 2002 by Preda Mihăilescu at Paderborn University.

没发一个Fields给他？

FoxMe 写了： 2023年 1月 27日 17:13 Catalan's conjecture (or Mihăilescu's theorem) is a theorem in number theory that was conjectured by the mathematician Eugène Charles Catalan in 1844 and proven in 2002 by Preda Mihăilescu at Paderborn University.

哈哈，看来我看帖不仔细，我当时就看了眼catalan猜想的表述，没读维基上的其他内容。

但无论如何，能给出不依赖于catalan猜想的初等证明，也是索南的乐趣所在。