If $x \equiv y \pmod{4}$ then $n^x \equiv n^y \pmod{10} x,y$ and $n$ are intergers.
I know that:
Since $x \equiv y \pmod{4}$ we have $4|x-y \Rightarrow n^4|n^{x-y}$.
$n^x-n^y = n^y( n^{x-y}-1)$
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Bill Dubuque
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1Note that $\phi(10)=4$. – Dietrich Burde Aug 26 '23 at 16:50
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1See the Theorem & Corollary in the linked dupe (using $e=ϕ(m)$) & Euler's Theorem). Here $,m=10,$ so $,\phi(m) = 4\ \ $ – Bill Dubuque Aug 26 '23 at 17:01
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Certainly not true if $x=0,y=4, n=2$ It doesn't even make sense if $x$ or $y$ is negative. But true if $x,y$ are positive integers. – Thomas Andrews Aug 26 '23 at 17:07