Let $p$ denote an odd prime and let $k \geq 1$ be a positive integer.

(a) Prove that there is only one primitive Pythagorean triple $(a, b, c)$ with $a = p^k$.

(b) Prove that there are exactly $k$ (not necessarily primitive) Pythagorean triples of the form $(a, b, c)$, where $a = p^k$.


closed as off-topic by Zev Chonoles, Morgan Rodgers, Claude Leibovici, TheSimpliFire, The Phenotype Mar 6 '18 at 10:17

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    $\begingroup$ Hint: If $a,b,c$ is a pythagorean triple then $a^2 = c^2 - b^2 = (c-b)(c+b)$. $\endgroup$ – fleablood Mar 6 '18 at 1:12

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

So to find a pythagorean triplet we can let $a$ be anything and $j = c-b; k = c+b$ can be any complimentary factors of $a^2$ provided $j,k$ are both the same parity.

So if $a=p^k$ where $p$ is prime, them $j = c-b$ and $k= c+b$ have only $p$ as prime factors. And if $p|c-b$ and $p|c+b$ then $p|(c-b)+(c+b)= 2c$ and $p|(c+b) - (c-b) = 2b$. If $p$ is odd then $p|c$ and $p|b$.

But $a,b,c$ are primitive so that is impossible. So the only option if $p$ is odd is that $c+b = p^{2k}$ and $(c-b)= 1$. Or in other words. $c = \frac {p^{2k} + 1}2$ and $b = \frac {p^{2k} - 1}2$ and $a^2 + b^2 = p^{2k} + \frac{p^{4k} - 2p^{2k} + 1}4 = \frac {p^{4k} + 2p^{2k} + 1}4 =(\frac {p^{2k} + 1}2)^2 = c^2$.

Now if $a,b,c$ are not primitive we are allowed for $p|b$ and $p|c$.

By the reasoning above we must have $p^{2k} = (c-b)(c+b) = j*k$ where $j = c-b = p^m$ and $k = c+b = p^l$ and $m + l = 2k$.

So $c = \frac {p^m + p^l}2$ and $b = \frac {p^m - p^l}2$. As $l < m$, $l$ may be anything for $0$ to $k-1$ and $m$ will be $2k - l$ which will spam $k+1$ to $2k$.

So there are precisely $k$ such triplets.


Since $p$ is an odd prime, $p^k$ is odd as well. By Pythagoras'rule, if $a$ is an odd number, then there exists $(a,\frac{a^2-1}{2},\frac{a^2+1}{2})$. Thus, your Pythagorean triple is: $(p^k,\frac{P^{2k}-1}{2},\frac{p^{2k}+1}{2})$ and it is primitive.

PS: Notes for Fleabood's proof:

By Fleabood's reasoning above, since we have $j = c-b$ and $k = c+b $, there should be two natural numbers $m$ and $l$ such that $m + l = 2k$, hence $p^{2k} = (c-b)(c+b) = jk$ where $j = c-b = p^m$ and $k = c+b = p^l$.

Thus, there exists a series of Pythagorean triples $(a,b,c)$ where $a=p^k$, $b = \frac {p^m - p^l}2$ and $c = \frac {p^m + p^l}2$. As pointed above, if $l < m$, $l$ may be anything for $0$ to $k-1$ and $m$ will be $2k - l$, so there are precisely $k$ such triplets (see above).

Suppose $m=l+x$ since $m$ and $l$ are natural numbers and $m\gt l$. Thus, $p^m=p^{l+x}=p^lp^x$ and $p^{2k}=p^{l+m}=p^{l+l+x}=p^{2l+x}=p^{2l}p^x$. When substitute these values for Pythagorean triples $(a,b,c)$, we have: $$a^2+b^2=c^2$$ $$p^{2k}+\left(\frac {p^m - p^l}2\right)^2=\left(\frac {p^m + p^l}2\right)^2$$ $$p^{2l}p^x+\left(\frac {p^lp^x - p^l}2\right)^2=\left(\frac {p^lp^x + p^l}2\right)^2$$ $$p^{2l}p^x+p^{2l}\left(\frac {p^x - 1}2\right)^2=p^{2l}\left(\frac {p^x + 1}2\right)^2$$ Thus, these Pythagorean triples are not primitive. Overall, it proves that only one primitive Pythagorean triple, $(p^k,\frac{P^{2k}-1}{2},\frac{p^{2k}+1}{2})$, exists with $a=p^k$ (the odd leg). The rest of the triples, $(p^k,\frac{P^m-p^l}{2},\frac{p^m+p^l}{2})$,with $a=p^k$ in the series are non-primitive.

  • $\begingroup$ how does it proof that there's only 1? And how about the second one $\endgroup$ – Gareth Ma Mar 6 '18 at 1:46
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    $\begingroup$ @KGSH bteam Mine Team Beast O: Thanks for your interest. I apologize to take too much time to answer your question because I had to leave in hurry for other commitments. That's the same reason why I left the question half answered. Well, meantime, Fleablood has very detailed answer, which would also be a good answer for your question. Let me know if you satisfied. $\endgroup$ – Mathew Mahindaratne Mar 6 '18 at 16:35

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