Quadratic congruences equivalent statements

Question

Prove that for every prime $p>3$,

$L= x^2-x+2 \equiv 0\mod p$ has a solution iff $D=x^2-x+16 \equiv 0\mod p$ has a solution.

This is not true, right? If $L \equiv 0$ mod$p$, then $D=L+14 \equiv 14\mod p$, and $14$ is not congruent to $0\mod p$, when $p \neq 7$.

You are assuming that the $x$ that solves the first congruence is the same as the $x$ that solves the second. — Angina Seng, Jan 24 '19 at 17:50
Duplicate of Let $p$ be a prime number. Prove that there exists $x∈Z$ for which $p|x^2−x+ 3$ if and only if there exists $y∈Z$ for which $p|y^2−y+ 25$. — Bill Dubuque, Jan 24 '19 at 19:17
@BillDubuque Can you give me a link on something that explains connection between discriminant and quadratic congruences ? — , Jan 24 '19 at 19:53
@someone Completing the square is universal, i.e. it works over any commutative ring, e.g. see here. Seel also here — Bill Dubuque, Jan 24 '19 at 20:12

score 0 · Answer 1 · edited Jan 25 '19 at 02:49

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$$(2x-1)^2\equiv-7\pmod p$$

$$(2y-1)^2\equiv-63\equiv\{3(2x-1)\}^2$$

edited Jan 25 '19 at 02:49

Bill Dubuque

answered Jan 24 '19 at 17:53

lab bhattacharjee

score 0 · Answer 2 · answered Jan 24 '19 at 17:56

0

Hint:

The discriminant of $x^2-x+2$ is $-7$.

The discriminant of $x^2-x+16$ is $-63=3^2(-7)$.

answered Jan 24 '19 at 17:56

lhf

score 0 · Accepted Answer · edited Jan 24 '19 at 18:29

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The first congruence has solutions if and only if $\Delta_1=-7$ is a square$\bmod p$.

The second congruence has solutions if and only if $\Delta_2=-63$ is a square$\bmod p$.

Now it is obvious that $$-7 \,\text{ is a square}\bmod p\iff-63=3^2\cdot -7 \,\text{ is a square}\bmod p.$$

edited Jan 24 '19 at 18:29

Angina Seng

answered Jan 24 '19 at 18:00

Bernard

3 Answers3