Show that one of these numbers is a Square

Question

For any $a,b\in\mathbb{Z}$, show that at least one of the following values is a square in $\mathbb{Z}_p$ with $p$ prime

Those values are:

$a^2 -b^2$, $2(b-a)$, $2(-a-b)$.

I tried multiplying, adding and substracting those values but I can't conclude anything.

edit: I forgot to mention that the numbers must be squares mod $p$

It seems none of them is necessarily the square of an integer. — Quillo, Jun 16 '20 at 00:15
I think the real question concerned quadratic residues modulo some odd prime. — Will Jagy, Jun 16 '20 at 00:15
Well if $b= 3$ and $a=2$ then $a^2 - b^2 = -5$ and $2(b-a)= 2$ and $2(-a-b) =-10$ and none of those are squares. I'm a bit perplex why it uses $2(-a-b)$ which is negative if $a,b$ are positive. — fleablood, Jun 16 '20 at 00:15
@fleablood the question was edited about five minutes after your comment — Will Jagy, Jun 16 '20 at 00:55
@WillJagy the post is still incorrect because none of them are squares. — Radial Arm Saw, Jun 16 '20 at 02:21
@RadialArmSaw By editing to be about modulo arithemetic mod a prime $p$ it is corrected. For example my example of $b=3$ and $a=2$ holds for $\mathbb Z_7$ because $a^2-b^2\equiv -5\equiv 2\equiv 9 \equiv 3^2 \pmod 7$. And for $\mathbb Z_{17}$ then $2(b-a)\equiv 2 \equiv 36 \equiv 6^2 \pmod {17}$ and so on. — fleablood, Jun 16 '20 at 02:54

Thomas Andrews · Answer 1 · 2020-06-16T02:48:13.673

4

Hint: $$a^2-b^2=(b-a)(-a-b)$$

Now, assume $a^2-b^2$ is not a square in $\mathbb Z_p.$

More generally, if $u,v,w$ are integers, then at least one of $uv,uw,vw$ is a square modulo $p.$ In your case, $u=b-a, v=-a-b, w=2.$

Even more generally, if $xyz$ is a square modulo $p,$ then so is at least one of $x,y,z.$

edited Jun 16 '20 at 02:48

answered Jun 16 '20 at 01:16

Thomas Andrews

177,126

1

Then either $b-a$ or $-a-b$ is not a square – Javier Vásquez Jun 16 '20 at 01:22
More importantly, one is a square and the other is not. @JavierVásquez – Thomas Andrews Jun 16 '20 at 01:53
@ThomasAndrews not when a=1 and b=4. In that case $b-a=3$ and $-a-b=-5$. Neither of those are squares. – Radial Arm Saw Jun 16 '20 at 02:19
Which $p$ are you using, @RadialArmSaw ? – Thomas Andrews Jun 16 '20 at 02:20
I don’t understand. I thought a “square” is a number that has a square root which is a whole number. – Radial Arm Saw Jun 16 '20 at 02:24
No, this is about squares modulo a prime $p.$ For example, when $p=7,$ $-5\equiv 3^2\pmod 7.$ @RadialArmSaw – Thomas Andrews Jun 16 '20 at 02:27
In your general case, when $p=7$ and $u,v,w=1,2,3$ is such a counter example. – W. Wongcharoenbhorn Jun 16 '20 at 02:42
1

No, $uv=2\equiv 3^2\pmod 7.$ @W.Wongcharoenbhorn – Thomas Andrews Jun 16 '20 at 02:44
"More generally, if u,v,w are integers, then at least one of uv,uw,vw is a square modulo p" That's a heck of a claim isn't it. Is that supposed to be an known result. – fleablood Jun 16 '20 at 02:49
Oh, great, my bad. What an interesting problem. – W. Wongcharoenbhorn Jun 16 '20 at 02:49
@fleablood It follows simply from the fact that the product of three non-squares, modulo $p$ is not a square modulo $p.$ Since $(uv)(uw)(vw)=(uvw)^2$ is a square, one of the three must be a square. – Thomas Andrews Jun 16 '20 at 02:52

Javier Vásquez · Answer 2 · 2020-06-16T04:59:09.517

0

Thanks for all of your comments. Let $p$ odd. Since $x\mapsto x^2$ induces a multiplicative endomorphism of $\mathbb{Z}_p^*$, and its image is the subgroup $F$ of square numbers, ir follows that the index $[\mathbb{Z}_p^* : F] =2$. Hence if $x,y,z$ are nonsquares in $\mathbb{Z}_p^*$,

$$(xF)(yF)(zF) = zF$$

Hence the product of three nonsquares is a nonsquare and we have proved the Thomas assertion

edited Jun 16 '20 at 04:59

answered Jun 16 '20 at 04:49

Javier Vásquez

67

Show that one of these numbers is a Square

2 Answers2