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Consider the following theorem:

If $(x,y,z)$ are the lengths of a Primitive Pythagorean triangle, then $$x = r^2-s^2$$ $$y = 2rs$$ $$z = r^2+z^2$$ where $\gcd(r,s) = 1$ and $r,s$ are of opposite parity.

According to the previous theorem,My try is the following:

since $x = r^2-s^2$, $x$ is difference of two squares implying that $x \equiv 0 \pmod 4$. But $x=21 \not \equiv 0 \pmod 4$. Hence, there are no triangles having such $x$.

Is that right?

Added:

My argument is false here. Please refer to the appropriate answer.

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    $\begingroup$ I don't understand your argument at all. It is simply not true that the difference of two squares is always $\equiv 0 \pmod 4$. $\endgroup$ – lulu Dec 15 '18 at 0:10
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    $\begingroup$ It was understandable before, but it is wrong. There are primitive triples with $21$. $(21,20,29)$, say $\endgroup$ – lulu Dec 15 '18 at 0:18
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    $\begingroup$ The hint I wrote out in an earlier comment is close to a complete solution. You should be able to follow it to list all the triples with $21$. $\endgroup$ – lulu Dec 15 '18 at 0:20
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    $\begingroup$ Squares are either $0 \text{ mod } 4$ or $1 \text{ mod } 4$, and hence there is at least one square which is $1 \text{ mod } 4$ and another which is $0 \text{ mod } 4$, whose difference will be $1 \text{ mod } 4$. $\endgroup$ – AlexanderJ93 Dec 15 '18 at 0:20
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    $\begingroup$ @MagedSaeed Do not mind for misakes, that's the way we learn a lot! Your question was fine and properly posted. Bye $\endgroup$ – gimusi Dec 15 '18 at 0:33
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Recall that

$$3^2+4^2=5^2 \implies (3\cdot 7)^2+(4\cdot 7)^2=(5\cdot 7)^2$$

and note that

$$(21, 220, 221)$$

is a primitive triple.

Your criterion doesn't works because the remainder of squares $\pmod 4$ are $0,1$ therefore we can't comclude that

$$z^2-y^2\equiv 0 \pmod 4$$

What we need to solve is

$$21^2=441=3^2\cdot 7^2=(z+y)(z-y)$$

that is we need to try with

  • $z-y=1 \quad z+y=441\implies (x,y,z)=(21,200,221)$
  • $z-y=3 \quad z+y=147\implies (x,y,z)=(21,72,75)$
  • $z-y=7 \quad z+y=63\implies (x,y,z)=(21,28,35)$
  • $z-y=9 \quad z+y=49\implies (x,y,z)=(21,20,29)$
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  • $\begingroup$ My method works if the question asks for primitive triangle. Right? $\endgroup$ – Maged Saeed Dec 15 '18 at 0:10
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    $\begingroup$ @MagedSaeed Note that also $(21, 220, 221)$ is a primitive triple. $\endgroup$ – gimusi Dec 15 '18 at 0:13
  • $\begingroup$ Can you find a general form of the solutions please. Refer to the question title. $\endgroup$ – Maged Saeed Dec 15 '18 at 0:18
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    $\begingroup$ @MagedSaeed Your criterion doesn't work since we can have $z\equiv 1 \pmod 4$ and $y\equiv 0 \pmod 4$. $\endgroup$ – gimusi Dec 15 '18 at 0:20
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    $\begingroup$ @MagedSaeed I've added something more! You are welcome, Thanks Bye $\endgroup$ – gimusi Dec 15 '18 at 0:31
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We have $21=x=k(m^2-n^2),\, y=2kmn,\, z=k(m^2+n^2)$ where $m,n, k \in \Bbb N$ with $\gcd (m,n)=1$ and $m,n$ not both odd.

So $(m^2-n^2,k)\in \{(1,21),(3,7),(7,3),(21,1)\}.$ Now $m^2-n^2=1$ is impossible, so $(m,n,k)\in \{(2,1,7), (4,3,3),(11,10,1),(5,2,1)\},$ giving $$(x,y,z)\in \{ (21,28,35), (21,72, 75),(21,220, 221),(21, 20, 29)\}.$$ We have $m\leq 11$ because if $m\geq 12$ then $x\geq m^2-n^2\geq m^2-(m-1)^2=2m-1\geq 23>21...$ There are 2 solutions $(11,10)$ and $(5,2)$ to $m^2-n^2=21.$

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