True or false: If $x^2 \equiv 14 \pmod{15}$ then there exists an $x^2$ that makes this congruent

$0^2\equiv0 \mod 15$

$1^2\equiv1 \mod 15$

$2^2\equiv4 \mod 15$

$3^2\equiv9 \mod 15 $

$4^2\equiv1 \mod 15 $

$5^2\equiv10 \mod 15 $

$6^2\equiv6 \mod 15 $

I don't see this as a proof since there is no obvious pattern but right now my answer is going to be false.

  • $\begingroup$ You may also prefer considering this as $x^2\congr -1\mod 15$ $\endgroup$ – user285523 Dec 12 '15 at 5:35
  • $\begingroup$ @0.5772156649..., the $\LaTeX$ symbol for modular equivalence is $\equiv$ given by \equiv. $\endgroup$ – learner Dec 12 '15 at 7:06
  • $\begingroup$ Huh, good to know $\endgroup$ – user285523 Dec 12 '15 at 18:24

Suppose there is a solution. Now $15 | x^2-14$, which implies $3 | x^2 - 14$. So you have $x^2 = 14 = 2 (mod 3)$, which is not possible. Hence you have no solutions.


You were actually going the right way. If there is an integer solution $x$ to the congruence $x^2\equiv 14\pmod{15}$, then $\exists~k~(-7\leq k\leq 7)~\mid~x\equiv k\pmod{15}$.

All you have to do now is to prove the negation of the above statement, i.e., you have to show the following :

$$x\equiv k\pmod{15}\implies x^2\not\equiv 14\pmod{15}~\forall~k~(-7\leq k\leq 7)$$

Since $x^2\equiv (-x)^2$, it suffices to show that none of $0^2,1^2,2^2,\ldots,7^2$ are congruent to $14$ modulo $15$ to prove that $x^2\equiv 14\pmod{15}$ has no solutions.

You already did it till $6^2$. Just do one more ($7^2$) and you would have a complete proof. It's easy to check that $7^2=49\equiv 4\not\equiv 14\pmod{15}$.


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