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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 $a = p^k$.

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

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  • $\begingroup$ Welcome to stackexchange. You are more likely to get answers rather than downvotes or votes to close if you edit your question to show what you have tried and where you are stuck. Please use mathjax: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Ethan Bolker Mar 6 '18 at 1:07
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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
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$a^2 + b^2 = c^2 \implies a^2 = c^2 - b^2 = (c-b)(c+b)$

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.

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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.

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  • $\begingroup$ 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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