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How can I find all $(a,b,c) \in \mathbb{Z}^3$ such that $a^2+b^2+ab$, $a^2+c^2+ac$ and $b^2+c^2+bc$ are squares ?

Thanks !

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2 Answers 2

I have shown here that:

All coprime triples $(a,b,c)$ so that $a^2 + ab + b^2 = c^2$ can be enumerated, without duplication, by taking two positive integers $m \ge n$, where $3$ does not divide $n$, and either $mn$ is odd and $\gcd(m,n) = 1$, or $8$ divides $mn$ and $\gcd(m,n) = 2$, and by setting $$ \begin{align} a&=mn\tag{1a}\\[9pt] b&=\frac{(3m+n)(m-n)}{4}\tag{1b}\\[9pt] c&=\frac{3m^2+n^2}{4}\tag{1c} \end{align} $$

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Seriously, $>=$? What happened to $\geq$, or $\geqslant$? :-) –  Asaf Karagila Jul 5 '12 at 20:21
    
@AsafKaragila: So, I missed that in the translation from ASCII :-p –  robjohn Jul 5 '12 at 20:42
    
Quite the husky translation, I'd say! –  Asaf Karagila Jul 5 '12 at 20:45

Let $c^2+a^2+ca= (c+na)^2$ where $n$ is an integer $\implies a=\frac{(2n-1)c}{1-n^2}$.

Let $a^2+b^2+ab =(a+mb)^2$ where $m$ is an integer $\implies b=\frac{(2m-1)a}{1-m^2}=\frac{(2n-1)(2m-1)}{(1-n^2)(1-m^2)}c$.

If $c|(1-n^2)(1-m^2)$,

$c=r(1-n^2)(1-m^2)$ (say, where $r$ an integer),

then, $b = r(2n-1)(2m-1)$

and $a = r(2n-1)(1-m^2)$

The above will be true if only 1st two conditions were supplied.

For all the three conditions, let $c^2+a^2+ca= r^2$ where r is an integer =>$r^2+ca$ must be perfect square and vice versa. ca=$r^2-R^2$ for some integer R. So, the given problem is same as finding a,b,c such that the product of any two is the difference of two squares.

Now $r^2+ca$ will be perfect square if r=$\frac{c-a}{2}$ where c-a is odd i.e., c,a are of opposite parity.

Then, $c^2+a^2+ca=(\frac{c-a}{2})^2$=>c+a=0.

Similarly, b+c=a+b=0. This provides only trivial solution (0,0,0) $ \in \mathbb{Z}^3$.

As ca=$r^2-R^2$, bc = $s^2-S^2$ and ab = $t^2-T^2$ for some integer s,S,t,T.

So, $a^2=\frac{ca.ab}{bc} =\frac{(r^2-R^2).(s^2-S^2)}{t^2-T^2} $

Now if r=$A^2+B^2$ and R= $A^2-B^2$

and if s=$C^2+D^2$ and S= $C^2-D^2$

and if t=$E^2+F^2$ and T= $E^2-F^2$ for some integers A,B,C,D,E,F.

a=±$\frac{2.A.B.2.C.D}{2E.F}$=±$\frac{2A.B.C.D}{E.F}$

b and c with be of the form ±$\frac{2.C.D.E.F}{A.B}$ and ±$\frac{2.E.F.A.B}{C.D}$

So, we need to find integers p,q,r such that p|2qr, q|2rp and r|2pq (think p=A.B, q=C.D and r=E.F).

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I am a bit baffled by this. Take the case $5^2+3^2+5\times 3 = 7^2$ (in other words $c=5$, $a=3$), then the first line above would give $(c+na)^2 = 7^2$, or $5+3a = 7$, and $n$ is then not an integer. Am I being a bit dim (as often the case!) –  Old John Jul 5 '12 at 18:35
    
@OldJohn In this particular case we can take $n=-4$. I do share your skepticism if $a$ has more than one odd prime factor (this seems to be assuming that $c^2$ has only two square roots mod $a$). Also it doesn't appear that $b^2+c^2+bc$ has been addressed. –  Erick Wong Jul 5 '12 at 18:51
    
OK, but is the solution above claiming that for any choice of $r$, $m$, $n$ in $\mathbb{Z}$, the last the lines will give a solution? (My calculations show otherwise - probably agreeing with your doubts about $b^2+c^2+bc$.) –  Old John Jul 5 '12 at 18:55

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