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Let $z>y>x$ primitive pythagorean triple. Show that $z$ has a prime factorization to primes of the form $4k+1$

I tried using the fact that there exist $s,t$ such that $s$ is not congruent to $t$ modulo $2$ and $z=s^2+t^2$ but it could only get me to the point where I see that $z$ itself is of the form $4k+1$

Bill Dubuque
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Tamir Vered
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  • @DietrichBurde, the question that you refer to tries to show that z itself is not of the form 4k+3 not that all the primes in it's factorization are, as I said I was able to show that z itself is 4k+1 – Tamir Vered Jul 03 '23 at 17:50
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    If $a^2+b^2\equiv 0\pmod p$ for a prime of the form $4k+3$, then $a\equiv b\equiv 0\pmod p$. – lulu Jul 03 '23 at 17:55
  • can you please show why? – Tamir Vered Jul 03 '23 at 22:38
  • If not you could make a square root of $-1$. – lulu Jul 03 '23 at 22:56
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    here are several more detailed answers to the $a^2+b^2\equiv 0 \pmod p$ question. Easiest is to note that, were $b$, say, non $0$ then $\left (\frac ab\right)^2\equiv -1\pmod p$. But there are other approaches. – lulu Jul 04 '23 at 10:23

2 Answers2

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If $a^2+b^2=c^2$ and this is a Pythagorean triple, then as mentioned, both $a$ and $b$ can't be even. If they are both odd then $a=2a'+1$, and $b=2b'+1$, $4a'^2+4a'+4b'^2+4b'+2=c^2$. But $c$ has to be even so $c=2c\implies c^2=4c'^2$. Contradiction, so one of $a$ and $b$ is even and the other odd.

WLOG $a$ is even and $b$ is odd. Then $c$ is odd.

$a^2=c^2-b^2=(c-b)(c+b)$

$a=2a', \ 4a'^2=(c-b)(c+b)$

$a'^2=\frac{1}{2}(c-b)\frac{1}{2}(c+b)$

$gcd(c-b,c+b)|2c$ and $gcd(c-b,c+b)|2b$. This tells us that $c-b$ and $c+b$ are relatively prime, apart from the factor of 2 since were they to have any other factor this would imply $gcd(b,c)>1$ contradicting these forming a Pythatogrean Triple.

It further follows that $(c-b)/2$ and $(c+b)/2$ are perfect squares.

Let $\frac{1}{2}(c+b)=n^2$ and $\frac{1}{2}(c-b)=m^2$

Then $c=m^2+n^2$. $b=n^2=m^2$.

$c^2-b^2=m^4+n^4+2m^2n^2-(m^4+n^4-2m^2n^2)=4m^2n^2=a^2$

So $a=2mn$.

All told, we see that the hypotenuse, $c$ is the sum of two squares.

The square of a number is congruent to 0 or 1 modulo 4. It follows that the sum of two squares is always $0,1,$ or $2$ moduloe $4$, but never 3.

So $c$ is odd and never congruent to $3$ modulo $4$. It follows that $c$ is always congruent to $1$ modulo $4$.

Suppose $x=r^2+s^2$ and $y=p^2+q^2$.

Then $xy=(r^2+s^2)(p^2+q^2)= (r+is)(r-is)(p+iq)(p-iq)=(r+is)(p+iq)(r-is)(p-iq)$

And $xy=[(rp-sq)+i(rq+ps)][(rp-sq)-i(rq+ps)]=(rp-sq)^2+(rq+ps)^2$

In other words, the product of two integers each expressible as the sum of two perfect is itself expressible as the sum of two perfect squares.

So if $p\equiv 3 \pmod {4}$ and $p|c$, then $p^2|c$

$p^2|c \land p^2|m^2\implies p^2|n^2\implies gcd(b,c)>1$, a contradiction.

$m^2+n^2\equiv 0 \pmod{p^2}$

GAUSSIAN INTEGERS. TBC

TurlocTheRed
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In the case of $\,z=s^2+t^2,\,$ the $\,z$-value is the square of a hypotenuse. The smallest primitive hypotenuse values of Pythagorean triples are shown here. Two formulas for generating these are:

$$\sqrt{z}=m^2+k^2\quad\text{and}\quad \sqrt{z}=(2n-1)^2+2(2n-1)k+2k^2$$

For the first formula to work, $\,(m,k)\,$ must be of opposite parity. For the second formula, any pair of natural numbers will generate a non-trivial Pythagorean triple.

The case of $\,(m,k)=(2,1) \implies \sqrt{z}=4+1=5=1\cdot5.$

The case of $\,(n,k)=(1,1) \implies\sqrt{z}=1+2+2=5==1\cdot5.$

The case of $\,(m,k)=(7,4) \implies \sqrt{z}=49+16=65=1\cdot5\cdot13.$

The case of $\,(m,k)=(8,1) \implies \sqrt{z}=64+1=65=1\cdot5\cdot13.$

The case of $\,(n,k)=(2,4) \implies\sqrt{z}=9+24+32=65=1\cdot5\cdot13.$

No matter what integers are input, the result will always factor into primes of the form $\,4k+1.$

poetasis
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