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Problem 2040

National Olympiad second round; IMO P1/P4
Number theory Difficulty 7.5 Prove it IMO Hk TST · Hong Kong

Let nn be a positive integer. Show that if pp is a prime dividing 54n−53n+52n−5n+15^{4n} - 5^{3n} + 5^{2n} - 5^n + 1, then p≡1(mod4)p \equiv 1 \pmod 4.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Clearly, p≠2,5p \neq 2, 5. Let m=54n−53n+52n−5n+1m = 5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1. Then
(2⋅52n−5n+2)2−5⋅52n=4m≡0(modp). (2 \cdot 5^{2n} - 5^n + 2)^2 - 5 \cdot 5^{2n} = 4m \equiv 0 \pmod{p}.
This gives 5≡(5−n(2⋅52n−5n+2))2(modp)5 \equiv (5^{-n}(2 \cdot 5^{2n} - 5^n + 2))^2 \pmod{p}. Using the Legendre symbol, we have (5p)=1\left(\frac{5}{p}\right) = 1. On the other hand, we have
(52n−5n+1)2+5n(5n−1)2=m≡0(modp). (5^{2n} - 5^n + 1)^2 + 5^n(5^n - 1)^2 = m \equiv 0 \pmod{p}.
As above, this implies (−5np)=1\left(\frac{-5^n}{p}\right) = 1. It follows that
(−1p)=(−5np)(5np)=1⋅1n=1. \left(\frac{-1}{p}\right) = \left(\frac{-5^n}{p}\right) \left(\frac{5^n}{p}\right) = 1 \cdot 1^n = 1.
Therefore, p≡1(mod4)p \equiv 1 \pmod 4.

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