I believe this is to do with modular arithmetic, maybe along the lines of factorising $2020$ into primes and working with totients .
$2020= 2 \cdot 2 \cdot 5 \cdot 101$.
The Euler totients of $4,101,$ and $5$ are $2,100$ and $4$ respectively. I am unsure of how to proceed further.
Any help would be appreciated.
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3I suggest you edit your question to show the factorization of $2020$ and then consider the totient of the resulting prime powers. Effectively we are a small step beyond Euler's theorem. – Joffan Mar 29 '21 at 14:48
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$,2^2\cdot 5\cdot 101\mid n^2(n^{100}-1),$ by $,e=2,f=100,$ in the simple Theorem in the linked dupe. – Bill Dubuque Mar 29 '21 at 17:44
2 Answers
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$2020=4×5×101$ which factors are pairwise relatively prime.
Prove the given function is a multiple of $4$. Use the fact that squares are all in $\{0,1\}\bmod 4$.
Prove it's a multiple of $5$. Use Fermat's Little Theorem which asserts $n^p\equiv n\bmod p$ for prime $p$.
Prive it's a multiple of $101$. See (2).
Oscar Lanzi
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Rewrite as:
$$x^{102}-x^2=x(x^{101}-x)$$
Due to Fermat little theorem:
$$x^{101}-x\equiv 0 \bmod 101$$
Rewrite as:
$$x^{102}-x^2=x^2[(x^{25})^4-1]$$
$(x^{25})^4-1\equiv 0 \bmod 5$
$\phi(4)=2$
$\Rightarrow (x^{50})^2-1\equiv 0\bmod 4$
$\Rightarrow x^{100}-1\equiv \bmod 20$
Finally:
$$x^{102}-x^2\equiv 0\bmod (4\times 5\times 101=2020)$$
sirous
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