Solve in positive integers $x^{100}-y^{100}=100!$
Problem
Source: Serbia JBMO TST 2020 P2
Tags: number theory
InternetPerson10
05.09.2020 09:52
Hint Show first that $x, y > 100$. Then bound to finish.
I claim that there are no solutions. It's easy to see that all the solutions I wrote work.
Clearly $x > y$. We first show that $\gcd(x, y) = 1$; assume otherwise for the sake of contradiction. Note that $$\nu_p(100!) = \left\lfloor \dfrac{100}{p} \right\rfloor + \left\lfloor \dfrac{100}{p^2} \right\rfloor + \left\lfloor \dfrac{100}{p^3} \right\rfloor + \cdots \leq \left\lfloor \dfrac{100}{2} \right\rfloor + \left\lfloor \dfrac{100}{2^2} \right\rfloor + \left\lfloor \dfrac{100}{2^3} \right\rfloor + \cdots = 50 + 25 + 12 + 6 + 3 + 1 = 97 < 100$$Consider a prime $p$ dividing $\gcd(x, y)$. Then $\nu_p(x^{100} - y^{100}) \geq 100$, contradiction. Thus $\gcd(x, y) = 1$.
Now, take a prime $p < 100$ that divides only one of $(x, y)$. Then $\nu_p(x^{100} - y^{100}) = 0 < \nu_p(100!)$, contradiction. Therefore no prime less than 100 can divide either $x$ or $y$. This implies $x, y > 100$. But $$101^{100} - 100^{100} = \binom{100}{1}100^{99} + \binom{100}{2}100^{98} + \binom{100}{3}100^{97} + \cdots > 100^{100} > 100!,$$contradiction! The value of $x^{100}-y^{100}$ would always be bigger than $100!$ for positive integers $x > y$. Thus there are no solutions. Yay! :DD
P1kachu
05.09.2020 10:03
Easy to see that $x>y$, and that $x$ and $y$ have same parities. Also, $\nu_2(100!)=97$.
Case I: $x$ and $y$ are even
$$\nu_2(x^{100}-y^{100})\ge100$$which is a contradiction as $\nu_2(100!)=97$.
Case II: $x$ and $y$ are odd
By LTE,
$$\nu_2(x^{100}-y^{100})=\nu_2(x-y)+\nu_2(x+y)+\nu_2(100)-1=\nu_2(x-y)+\nu(x+y)+1$$and since $\nu_2(100!)=97$, this means $\nu_2(x-y)+\nu_2(x+y)=96$.
Since $\gcd(x-y,x+y)=\gcd(x-y,2y)$ and $y$ is odd, this means either $\nu_2(x-y)$ or $\nu_2(x+y)$ must equal 1. This gives two cases:
If $\nu_2(x-y)=1$, then $\nu_2(x+y)=95$, which imply $x+y\ge2^{95}\implies x\ge2^{94}$.
If $\nu_2(x+y)=1$, then $\nu_2(x-y)=95$, which imply $x-y\ge2^{95}\implies x\ge2^{95}$.
In both cases, $x\ge2^{94}$. Observe that
$$x^{100}-y^{100}\ge x^{100}-(x-1)^{100}\ge 2^{94\cdot 100}-(2^{94}-1)^{100}>100\cdot(2^{94}-1)^{99}>100!$$which is a contradiction.
Steve12345
05.09.2020 12:03
FLT: $x^{100} \equiv y^{100} \equiv 1 (mod \ 101)$ (101 is prime), if $101$ doesn't divide $x$ nor $y$. $1)$ if $101|x,y$ obviously doesn't work. $2)$ if $101|y$ and doesn't divide $x$, by Wilsons this can't happen $3)$ if $101|x$ and doesn't divide $y$, so $x=101k$, but here we will have minimal difference if $y=100$ ($x$ and $y$ aren't equal) and $k=1$ (everything else will throw it in the negatives or make a bigger gap), but $101^{100}-100^{100}=(101-100)(101^{99}+...+100^{99})> 101^{99}+100^{99} > 2*100^{99} = 100 \cdot 100\cdot ... \cdot 100 \cdot 2 > 100 \cdot 99 \cdot 98 \cdot ... \cdot 2 = 100!$
sttsmet
17.06.2021 09:05
By Wilson Theorem you mean that for every $p \to (p-1)! \equiv -1 ( mod \ p )$ where p is a prime ???
Assassino9931
25.01.2022 20:47
If $x$ and $y$ are both not divisible by $101$, then $x^{100} \equiv y^{100} \pmod {101}$ by Fermat's Little Theorem, so $100!$ would again be divisible by $101$, impossible. If $x$ is not divisible by $101$ but $y$ is, then the left hand side would give $1$ by Fermat, while $100! \equiv -1 \pmod {101}$ by Wilson's theorem, contradiction. So we may assume that $x$ is divisible by $101$. But then $x\geq 101$ and since $x>y$ we have $y\leq x-1$, so $2\cdot 3 \cdots 99 \cdot 100 = 100! = x^{100} - y^{100} \geq x^{100} - (x-1)^{100} = (x - (x-1))(x^{99} + x^{98}(x-1) + \cdots + x(x-1)^{98} + (x-1)^{99}) > 101^{99}$, contradiction. Therefore there are no solutions.