A fourth-power congruence without dividing by 2 modulo n

mana_snack·Mar 17, 2026

Newton identities look useful, but n can be even, so dividing by 2 seems risky. Can we prove the fourth-power statement with integer identities only? Problem: Let

n

be a positive integer and let

a, b, c, d

be integers such that

n|a + b + c + d

and

n|a^2 + b^2 + c^2 + d^2

. Show that

n|a^4 + b^4 + c^4 + d^4 + 4abcd.

Source: Baltic Way 2016 · MNET-01DS

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2 replies

link_in_bio_·Mar 17, 2026

Let

w(t)=(t-a)(t-b)(t-c)(t-d)=t^4+At^3+Bt^2+Ct+D

. Sum

w(a)=w(b)=w(c)=w(d)=0

. This gives

p_4+Ap_3+Bp_2+Cp_1+4D=0

, where

p_j=a^j+b^j+c^j+d^j

mana_snack·Mar 18, 2026

Since A=−p₁ and D=abcd, the desired expression is

p_1p_3-Bp_2-Cp_1

, a sum of multiples of n. No division at all.

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