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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 $y^{100}=1 \pmod {101}$

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    $\begingroup$ Please show your attempts. $\endgroup$ – Student1058 Sep 3 '20 at 10:10
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    $\begingroup$ 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 $y$ ?... $\endgroup$ – TheSilverDoe Sep 3 '20 at 10:10
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    $\begingroup$ @Philip do you mind putting your attempts along with the question. This will avoid downvotes! $\endgroup$ – Albus Dumbledore Sep 3 '20 at 10:47
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    $\begingroup$ @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 :-) $\endgroup$ – Jyrki Lahtonen Sep 3 '20 at 10:52
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    $\begingroup$ @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 ! $\endgroup$ – TheSilverDoe Sep 3 '20 at 10:55
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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 $100>1,2,3,\cdots,99$.

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  • $\begingroup$ How could I overlook this method with the difference of consecutive $100$ th powers (+1) $\endgroup$ – Peter Sep 3 '20 at 11:42
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    $\begingroup$ 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. $\endgroup$ – Peter Sep 3 '20 at 11:47
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    $\begingroup$ @tkf This is not what I mean : $100\cdot 100^{99}=100^{100}$ is what I noticed after the comment of TheSilverDoe $\endgroup$ – Peter Sep 3 '20 at 12:11
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    $\begingroup$ @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? $\endgroup$ – tkf Sep 3 '20 at 12:33
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    $\begingroup$ Fine now, I hope the author will accept it. $\endgroup$ – Peter Sep 3 '20 at 12:43
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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.

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    $\begingroup$ 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. $\endgroup$ – user3733558 Mar 27 at 23:49
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    $\begingroup$ Oh, okay thanks. I'm new to the site. $\endgroup$ – Mathology Mar 28 at 0:07
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    $\begingroup$ 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. $\endgroup$ – user3733558 Mar 28 at 0:16
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    $\begingroup$ 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. $\endgroup$ – Mathology Mar 28 at 0:29
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    $\begingroup$ 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. $\endgroup$ – Mohsen Shahriari Mar 28 at 0:37

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