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 October 12th, 2015, 06:31 PM #1 Senior Member   Joined: Dec 2006 Posts: 166 Thanks: 3 Special Pythagorean triple? Is there a Pythagorean triple $(a,b,c)$ such that $a^2 -c$ and $b^2 -c$ are both perfect squares?
 October 12th, 2015, 08:08 PM #2 Senior Member   Joined: Dec 2006 Posts: 166 Thanks: 3 If $\gcd(a,b,c) = 1$, then $\{a,b\} = \{m^2 -n^2, 2mn\},\;c = m^2+n^2$ for some $m,n\in\mathbb{N}^+,\;\gcd(m,n) =1.$ Thus $(2mn)^2 -(m^2+n^2) = 4m^2n^2 -m^2 -n^2\equiv j\mod 4$ $j = \begin{cases}2, & (2\nmid m)\wedge(2\nmid n);\\ 3. & otherwise.\end{cases}$ and so one of $a^2 -c,\; b^2 -c$ is not perfect square. Therefore none of prime Pythagorean triples are the solutions of Original Problem.
 October 13th, 2015, 05:13 AM #3 Senior Member   Joined: Dec 2007 Posts: 687 Thanks: 47 Hi Elim, we could also use another direction I think is more or less what you are saying. For $c^2=a^2+b^2$, it is assumed that $c$ is odd if $(a,b,c)$ is primitive and $a,b$ have different parity. If both $a^2-c=k^2$ and $b^2-c=k'^2$ then $a^2+b^2=k^2+k'^2+2c=c^2$. It follows that $k^2+k'^2=c(c-2)$, put $c=d+1$ and we have $k^2+k'^2=(d+1)(d-1)=d^2-1$, where $d$ is even. Now $k^2+k'^2+1=d^2\equiv 0\pmod 4$, so $k^2+k'^2\equiv -1\pmod 4$, contradicting the fact that the pair $(k,k')$ has different parity, i.e., in this case we should have $k^2+k'^2\equiv 1\pmod 4$. Now your question did not ask for only the primitive case, so extra work is: Assume $\gcd(a,b,c)=r>1$ and take WLOG $a^2-c=k^2$. Put $a=ra'$ and $c=rc'$, then $r(ra'^2-c')=k^2$ and $r\mid k$, but in this case $ra'^2-c'=rk'^2$ so $r\mid c'$, absurd.
 October 13th, 2015, 10:51 AM #4 Senior Member   Joined: Dec 2006 Posts: 166 Thanks: 3 Thanks a lot al-mahed, very good approach! Here is my rephrase of your sol.: Assume that there are positive integers $a,b,c,u,v\underset{\,}{\;}$such that $\qquad\quad a^2 +b^2 = c^2,\; a^2 -c = u^2,\; b^2 -c = v^2$. Case I: $2\nmid c.$ Thus $a,\underset{\,}{\;}b$ have different parity and so $\qquad\quad u^2 + v^2 + 2c = a^2 + b^2 = c^2,\underset{\,}{\,}$ $\qquad\quad u^2 + v^2 + 1 = (c-1)^2\equiv 0\pmod{4}\underset{\,}{,}$ i.e. $\qquad\quad u^2 + v^2 \equiv 3\pmod{4}\underset{\,}{\;}$which can never happen $\qquad\quad$to a sum of two perfect squares. Case II: $2\mid c.$ Thus $(a,b,c) = r(a\,',b\,',c\,'),\,2\mid r,\,2\nmid c\,'\underset{\,}{.}$ $\qquad\quad$so $r(ra\,' - c\,') = a^2 -c = u^2,\; 4\mid u^2,\; 2\mid c\,'$,absurd. Thanks from al-mahed
 October 13th, 2015, 11:51 AM #5 Senior Member   Joined: Dec 2007 Posts: 687 Thanks: 47 Thanks Elim, your solution is more elegant and economical. Thanks from elim
 October 14th, 2015, 10:19 PM #6 Senior Member   Joined: Dec 2006 Posts: 166 Thanks: 3 But sorry I found that $r\mid u^2\implies 4\mid u^2\not\Rightarrow 2\mid c\,'$
 October 17th, 2015, 02:54 PM #7 Senior Member   Joined: Dec 2007 Posts: 687 Thanks: 47 Hi Elim, your Case II has a small mistake, maybe a typo, in the exponent of $a'$, but most important is that you can indeed conclude that $r\mid u^2\implies 4\mid u^2$, recall that $r$ is even and that $u^2$ is also even and a square, so it is divisible by 4. No problem here. Since $r$ is even, so $a^2$ is also even, then it must be the case that so called $c'$ is even. The problem with this proof (and now I admit I didn't read carefully your proof) is that you didn't rule out all cases, since in Case II it is shown that $c'$ must be even, and you can have such cases. For instance, $r=2^{t}\cdot r'$ for $t>1$. I see also that I've made a silly mistake, I cannot jump to the conclusion that $ra'^2-c'=ru'^2$ and that $r\nmid c'$ as I did without further ado, so with a correction the proof would go in this way using your schemes: Let $a,b,c,u,v\in\mathbb{N}$ be such that $\qquad\quad a^2 +b^2 = c^2,\; a^2 -c = u^2,\; b^2 -c = v^2$. Case I: $(a,b,c)$ is primitive, then $c$ is odd, writing $$u^2 + v^2 + 2c = a^2 + b^2 = c^2$$ $$u^2 + v^2 + 1 = (c-1)^2\equiv 0\pmod{4}$$ $$u^2 + v^2 \equiv -1\pmod{4}$$ But that's not possible since $u^2 + v^2=(2n+1)^2+(2m)^2=4(n^2+m^2+n)+1$ (here also you can use that theorem of sum of two squares). Case II: $(a,b,c)$ is not primitive, then $(a,b,c)=r(a',b',c')$ where $(a',b',c')$ is primitive. In this case take $r(ra'^2-c')=u^2$ so that $r\mid u^2$. Since for all primes such that $p\mid r$ we have $p\mid u^2$, and $p\mid u^2\iff p^2\mid u^2$, it is immediate that $u^2=r^2u'^2$ (the "further ado" required), so $ra'^2-c'=ru'^2$ and we conclude that $r\mid c'$. Write $c'=rc''$ (more "further ado" ) and we have $a'^2-c''=u'^2$, but then we are reduced to a similar situation as in Case I (keep in mind that $c'$ is odd): $$u'^2 + v'^2 + 2c'' = a'^2 + b'^2 = c'^2=r^2c''^2$$ $$r^2c''^2-2c''\equiv 1\pmod 4$$ $$r^2(1)-2(\pm 1)\equiv 1\pmod 4$$ $$r^2\equiv\pm 1\pmod 4$$ absurd. Last edited by al-mahed; October 17th, 2015 at 02:58 PM.

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