一个中等难度题目

[（ヅ）]

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

[YWY]

3^2 - 2^3 = 1。还有没有别的，我就不知道了，大家接着找。

[verdelite]

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

[verdelite]

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。

[（ヅ）]

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

[YWY]

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

2^1-1^2=1

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。

[Caravel]

基本题目里面有素数的题目就不好搞，我看着就走

[YWY]

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.

[（ヅ）]

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.

[YWY]

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.

[FoxMe]

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。

[YWY]

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。

[FoxMe]

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.

[TheMatrix]

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？

[YWY]

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。