# Congruence, prime number, integer

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• Jun 8th 2011, 05:28 AM
Shurelia
Congruence, prime number, integer
let p be odd prime, and a,b be integers
1) prove that the congruence $\displaystyle {x}^{2}-ab\equiv (a-b)x(mod p)$ always have one solution
2) when does the above congruence have 2 solutions modulo p?
3) solve the congruence $\displaystyle 4{x}^{3}\equiv x(mod p)$
• Jun 8th 2011, 08:10 AM
TheEmptySet
Quote:

Originally Posted by Shurelia
let p be odd prime, and a,b be integers
1) prove that the congruence $\displaystyle {x}^{2}-ab\equiv (a-b)x(mod p)$ always have one solution
2) when does the above congruence have 2 solutions modulo p?
3) solve the congruence $\displaystyle 4{x}^{3}\equiv x(mod p)$

The wording for part one is strange to me.

Consider

$\displaystyle x^2-ab \equiv (a -b)x \mod{p} \iff x^2 +(b-a)x-ab \equiv 0 \mod{p}$

Since the integers mod p are a finite field they are an integral domain we have that

$\displaystyle (x+b)(x-a) \equiv 0 \mod{p}$

I cannot see any reason why [tex]a \equiv -b \mod{p}[\tex] So it looks like it can have two solutions.

Please clarify and see if this gets you started on the next problem.

For the last one just factor!