$(x^2-a)(x^2-b)(x^2-ab)$ ≡ 0 (mod p)$ is solvable.

Question

Let p be an odd prime and let a,b ∈ Z such that p ∤ a. Prove that the congruence $(x^2-a)(x^2-b)(x^2-ab)$ ≡ 0 (mod p) is always solvable.

Not sure where to begin here.

Bill Dubuque · Accepted Answer · 2024-02-26T18:37:29.593

3

Hint $\ $ mod $p,\ $ if $\,a,b\,$ are not quadratic residues then $\,ab\,$ is one, i.e. if $\,x^2-a,\,x^2-b\,$ have no roots, then $\,x^2-ab\,$ has a root.

edited Feb 26 '24 at 18:37

answered Apr 14 '14 at 00:44

Bill Dubuque

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is this a therorem? – User011123521 Apr 14 '14 at 00:51
Also, I do not believe I have this as a theorem to use, otherwise I would have to prove it while im not sure as how to do so. – User011123521 Apr 14 '14 at 01:39
@Jonathan This is one of the first theorems one learns when studying quadratic reciprocity. It is highly unlikely that this problem would be posed before that result was proven. – Bill Dubuque Apr 14 '14 at 01:50

$(x^2-a)(x^2-b)(x^2-ab)$ ≡ 0 (mod p)$ is solvable.

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