# Solving $x^{100}− y^{100 }= 100!$

Please some hint on how to solve in the set of natural numbers $$x^{100} − y^{100} = 100!$$ The question comes from the Serbian Junior Mathematical Olympiad 2020.

I have tried with Fermat`s theorem using $$x^{100}=x^{101-1}$$, same with $$y$$ and that $$x$$ is a multiple of $$101$$ and $$y^{100}=1 \pmod {101}$$

• Please show your attempts. – Student1058 Sep 3 '20 at 10:10
• For any $y \geq 0$, $x= \sqrt[100]{100!+y^{100}}$ is a solution... Or maybe you make some assumptions on who are $x$ and $y$ ?... – TheSilverDoe Sep 3 '20 at 10:10
• @Philip do you mind putting your attempts along with the question. This will avoid downvotes! – Albus Dumbledore Sep 3 '20 at 10:47
• @TheSilverDoe Admittedly the tag choices made by a new poster are often unreliable. But in my humble opinion the addition of the tag number-theory comes with at least a strong inference if not an implicit assumption that the variables are integers :-) – Jyrki Lahtonen Sep 3 '20 at 10:52
• @JyrkiLahtonen Well, in the first version, there was also the tag "linear-algebra"... Finding the real solutions of this equation is as much number-theory as finding the integer solutions is linear-algebra ! – TheSilverDoe Sep 3 '20 at 10:55

You have shown that a solution requires $$x=101k$$. Therefore $$x^{100}-y^{100}\geq101^{100}-100^{100}\geq100\times100^{99}=100^{100}>100!$$

The first inequality follows from $$(101k)^{100}-y^{100}$$ taking its smallest positive value for fixed $$k\in \mathbb{N}$$ when $$y=101k-1$$ and $$(101k)^{100}-(101k-1)^{100}$$ being an increasing function in $$k\in\mathbb{N}$$ (and negative when $$k=0$$).

The second inequality follows from the binomial expansion of $$(100+1)^{100}$$.

The final inequality follows from $$100>1,2,3,\cdots,99$$.

• How could I overlook this method with the difference of consecutive $100$ th powers (+1) – Peter Sep 3 '20 at 11:42
• Since there is no solution for $0\le y\le 10^8$ , which I checked by brute force (which was of course a huge overkill , since limit $100$ would have already be sufficient), there cannot be integer solutions at all. – Peter Sep 3 '20 at 11:47
• @tkf This is not what I mean : $100\cdot 100^{99}=100^{100}$ is what I noticed after the comment of TheSilverDoe – Peter Sep 3 '20 at 12:11
• @Peter If $y=0$ then $x^{100}-y^{100}=x^{100}\geq x^{100}-(x-1)^{100}$. Which bit of the argument does not hold? – tkf Sep 3 '20 at 12:33
• Fine now, I hope the author will accept it. – Peter Sep 3 '20 at 12:43

I have a clear solution. First, let's apply Fermat's little theorem and we have $$x ^ { 100 } , y ^ { 100 } \equiv 1 \pmod { 101 }$$. Then we can use Wilson's theorem and get $$100 ! \equiv - 1 \pmod { 101 }$$. So, the left-hand side of the equation ($$x ^ { 100 } - y ^ { 100 }$$) is a multiple of $$101$$, but the right-hand side ($$100 !$$) is not. Therefore, no solutions exist in the set of natural numbers.

• Hi and welcome to Math.SE. It would be preferable to use MathJax for mathematical expressions. Check out math.stackexchange.com/help/notation to get started if you're unfamiliar with it. – user3733558 Mar 27 at 23:49
• Oh, okay thanks. I'm new to the site. – Mathology Mar 28 at 0:07
• It's alright, we were all newbies at some point. Stick around long enough, and you will soon be the one giving friendly advice to newcomers. – user3733558 Mar 28 at 0:16
• Yeah, but I happen to run into another problem. Well this proof is only valid for when x and y are NOT a multiple of 101 (rule of Fermat's Little theorem). I have a proof for when both x and y are a multiple of 101 but I'm still working on the case where only one of them is a multiple of 101. – Mathology Mar 28 at 0:29
• Another tip for you as a newcomer: as multiple users may have put comments under a post, you may want to notify the user you are answering to. That can be done using an at sign followed by the username. Example: "@Mathology". Note that in each comment, the system doesn't allow you to notify more than a single user. Hope you have fun, learn a lot and help many others on MSE. – Mohsen Shahriari Mar 28 at 0:37