Show a,b,c congruences when $a^2+b^2=c^2$

Solution 1:

Hints:

$$\forall n\in\Bbb Z\;\;,\;\;\;n^2=0,1\pmod 3\Longrightarrow a^2+b^2=c^2=0,1\pmod 3\Longrightarrow \ldots$$

Try now to come up with something similar as above for $\,p=4,5\,$

For the last one (and for (c), too), remember that $\,n^2=0,1\pmod 4\,$ , so...

Solution 2:

(a)

(i)If $3\not\mid ab, a\equiv\pm1\pmod3,b\equiv\pm1\implies a^2+b^2\equiv1+1\equiv-1\pmod 3$

but $c\equiv\pm1,0\implies c^2\equiv1,0\pmod 3$

(ii)Alternatively, using this, $a=k(m^2-n^2),b=k(2mn),c=k(m^2+n^2)$ will generate all Pythagorean triples uniquely where m, n, and k are positive integers with m > n, m − n odd, and with m and n co-prime.

So, $ab=k2mn\cdot k(m^2-n^2)$

If $3\mid mn, 3\mid b$ we are done else $(mn,3)=1$

Using Fermat's Little Theorem,

$m^2\equiv1\pmod 3$ and $n^2\equiv1\implies 3\mid(m^2-n^2)\implies 3\mid a$

(b)

(i) If $5\not\mid ab, a\equiv\pm1,\pm2\pmod3 \implies a^2\equiv1,4\pmod 5$

Similarly, $b^2\equiv1,4\pmod 5$

Observe that $c\equiv0,\pm1,\pm2\pmod5\implies c^2\equiv0,1,4\pmod 5$

If $a\equiv\pm1,b\equiv\pm1, a^2+b^2\equiv2\pmod5$ which is not congruent to any square $\pmod 5$

If $a\equiv\pm2,b\equiv\pm2, a^2+b^2\equiv8\equiv3\pmod5$ which is not congruent to any square $\pmod 5$

So, either $a\equiv\pm2,b\equiv\pm1$ or $a\equiv\pm1,b\equiv\pm2$

(ii)In the alternate way, $a=k(m^2-n^2),b=k(2mn),c=k(m^2+n^2)$

So, $abc=k^32mn(m^4-n^4)$

If $5|m$ or $5|n$ we are done else $(mn,5)=1$

Using Fermat's Little Theorem,

$m^4\equiv1\pmod 5$ and $n^4\equiv1\implies 5\mid(m^4-n^4)\implies 5\mid(m^2-n^2)$ or $5\mid(m^2+n^2)$

(c)$a=k(m^2-n^2),b=k(2mn),c=k(m^2+n^2)$

As $m-n$ is odd, $2\mid mn\implies 4\mid b$

The last problem : Observe that the square of any number is $\equiv0,1,4\pmod 8$