Is X a square in $\Bbb Z$?

Question

I am learning about number theory and cryptography. I have a style of question that I have not seen before and I do not have any examples to go off of.

The question is:

Is 50 a square in $\Bbb Z^x_{71}$? Find the principal square root of it, if the answer is positive.

Is this question solved similarly to Modular Arithmetic - Find the Square Root?

Since 71 is prime it cannot be broken down to any number but itself and 1, how do you solve for squares?

$50+71=121$, a square. – Gerry Myerson Nov 05 '17 at 06:03 — Gerry Myerson, Nov 05 '17 at 06:03
@GerryMyerson Is it really that simple? – TheDude1142 Nov 05 '17 at 06:04 — TheDude1142, Nov 05 '17 at 06:04
In this case, yes. In general, no. – Gerry Myerson Nov 05 '17 at 06:04 — Gerry Myerson, Nov 05 '17 at 06:04

score 0 · Accepted Answer · answered Nov 05 '17 at 06:04

You need to consider the quadratic residues modulo $71$. So this means computing $1^2,\ 2^2, \ 3^2 \ ... 70^2$ modulo $71$. The set of numbers that this forms is the quadratic residue mod $71$. When doing so, you will find that $50$ is in this residue class.

An easy way to think of it is as follows. You are not finding square roots, you are finding square numbers and in turn what the roots of these square numbers are.

Check this link out for more information. https://en.wikipedia.org/wiki/Quadratic_residue

"An easy way to think of it is as follows. You are not finding square roots, you are finding square numbers and in turn what the roots of these square numbers are." This line made it click for me. Thank you very much. — TheDude1142, Nov 05 '17 at 06:11

score 0 · Answer 2 · answered Nov 05 '17 at 06:06

0

For $x^2 < 71, 50$ is not a square. So now we take $x^2 = 71 + 50$ and it turns out $11^2 = 121 = 50(\mod 71)$

answered Nov 05 '17 at 06:06

Mikhail Tikhonov

532

This also was very helpful, thank you. – TheDude1142 Nov 05 '17 at 06:12

Is X a square in $\Bbb Z$?

2 Answers2