If $x^2+y^2=z^2$ has a solution then $5$ divides $xyz$

Question

I tried to solve a question but I did not succeed yet...my question is about number theory. Here it is:

How can we show that if the equation $x^2+y^2=z^2$ has a solution then $5$ divides $xyz$?
Can we have a general method for solving questions like that?

I need the beginning of the solution.

I will be very glad if you help.

Thanks

http://www.cut-the-knot.org/pythagoras/pythTripleDiv.shtml and http://mathoverflow.net/questions/4318/proof-of-if-a2-b2-c2-then-abc-is-divisible-by-60 — lab bhattacharjee, Oct 21 '13 at 08:21
Another useful link : http://mathworld.wolfram.com/PythagoreanTriple.html — lab bhattacharjee, Oct 21 '13 at 09:59
@EpicGuy: all this talk about "divides" kinda implies that we're dealing with integers only (which we are)... — The Chaz 2.0, Dec 18 '13 at 04:58

score 4 · Accepted Answer · answered Oct 21 '13 at 08:12

If $5$ divides $x$ and/or $y,$ we are done

Else

Observe that $$(\pm1)^2\equiv1,(\pm2)^2\equiv4\pmod 5$$

If both $x,y\equiv1\pmod 5, x^2+y^2\equiv2\pmod 5\not\equiv z^2$

If both $x,y\equiv2\pmod 5, x^2+y^2\equiv8\pmod 5\equiv3 \not\equiv z^2$

If $x\equiv\pm1,y\equiv\pm2,x^2+y^2\equiv0\pmod 5$

Now prove that if a prime $p$ divide $a^2,p$ must divide $a$

Here $p=5$

score 1 · Answer 2 · edited Apr 13 '17 at 12:20

1

If you know about general solution then you can see we can write primitive solution as $x=2st , y=s^2-t^2 , z=s^2+t^2$ , if $5$ divides one of $s$ or $t$ then $5|x$ and we are done. If $5$ does not divide any of $s$ and $t$ then by Fermat's little theorem $5|s^4-1 , 5|t^4-1$, so that $5|s^4-t^4$ i.e. $5|yz$ , this proves the claim.

edited Apr 13 '17 at 12:20

Community

1

answered Oct 21 '13 at 08:19

Souvik Dey

8,297

1

I have added a stackexchange reference to your answer – lab bhattacharjee Oct 21 '13 at 08:23
@ lab bhattacharjee: Thank you. – Souvik Dey Oct 21 '13 at 08:24

If $x^2+y^2=z^2$ has a solution then $5$ divides $xyz$

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