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(x+y+z)^2 \equiv axyz mod p

Source: India tst 2002 p14

July 13, 2012
modular arithmeticnumber theory unsolvednumber theory

Problem Statement

Let pp be an odd prime and let aa be an integer not divisible by pp. Show that there are p2+1p^2+1 triples of integers (x,y,z)(x,y,z) with 0≤x,y,z<p0 \le x,y,z < p and such that (x+y+z)2≡axyz(modp)(x+y+z)^2 \equiv axyz \pmod p
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