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Prove or disprove the claim that there are integers $n,m, 0<n<m$ such that $m^2 +mn+n^2$ is a perfect square.

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You can weaken the restriction to $0<n\le m$ since if $m=n$ the expression is $3m^2$ which cannot be square. –  vadim123 Dec 15 '13 at 21:54
    
Is this a homework question? What work have you done on it so far? –  Hovercouch Dec 15 '13 at 21:55
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Hovercouch: This is sort of homework. The original was with m and n being integers. Then it's easy to prove it as m=-n or m or n being 0 yields a square number. But I was curious about this modified version. I thought it should be easy but maybe because I'm getting tired but I couldn't prove it so far. –  elang Dec 15 '13 at 22:02
    
I deleted a (wrong) proof I had to the contrary. I will put it here instead: View the expression as a polynomial in $\mathbb{Z}[m]$. If it is irreducible over $\mathbb{Q}$, clearly it cannot be factored as a perfect square. If we reduce modulo $2$, we find that $n$ is even, otherwise the polynomial is irreducible. Similarly, we can find the $m$ is even. We can then factor a $4$ out of the whole expression, and we must still get a perfect square (since $4$ is a perfect square itself). Let $m' = m/2$ and $n' = n/2$. We get that $m'^2 + m'n' + n'^2$ is also a perfect square. Repeat to infinity. –  Eric Thoma Dec 15 '13 at 22:10

2 Answers 2

up vote 3 down vote accepted

For instance, let $m=5$ and $n=3$.

Remark: Here is a beginning to a general analysis. We want $x^2+xy+y^2$ to be a perfect square. It is easier to multiply by $4$, and solve $4(x^2+xy+y^2)=z^2$. Complete the square. So we want $(2x+y)^2+3y^2=z^2$. More generally look for solutions of $u^2+3v^2=z^2$. To produce solutions, we take a reasonable $v^2$, and express $3v^2$ as a difference of squares.

Added: In a comment OP asked whether there are infinitely many (relatively prime) pairs $(m,n)$.

Let $n$ be odd. Note that $$\left(\frac{3n^2-1}{2}\right)^2 + 3n^2=\left(\frac{3n^2+1}{2}\right)^2.$$ So if $2m+n=\frac{3n^2-1}{2}$, then the pair $(m,n)$ should do the job. That gives $m=\frac{3n^2-2n-1}{4}$. (It is easy to check that $3n^2-2n-1$ is divisible by $4$).

This by no means produces all pairs $(m,n)$ with the desired property.

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Thanks! I was so convinced that there are no such numbers that I didn't even try to look for counterexamples. No wonder I couldn't prove it. :-) –  elang Dec 15 '13 at 22:07
    
How difficult would be to try to answer the question if there are infinite many m,n pairs that satisfy the claim? –  elang Dec 15 '13 at 22:10
    
I happen to have sketched an approach. Of course there are infinitely many in a trivial way, $5k,3k$. But there are also infinitely relatively prime pairs. –  André Nicolas Dec 15 '13 at 22:13
    
Thanks Andre! :-) –  elang Dec 15 '13 at 22:22
    
You are welcome. I have added an explicit way to produce infinitely many examples. For instance take $n=5$. We get $m=16$, and $m^2+mn+n^2=19^2$. –  André Nicolas Dec 15 '13 at 22:36

Hint $\rm\ \ A^2 + AB+ B^2 = (a^2+ab+b^2)^2\ $ for $\rm\ A = a^2-b^2,\,\ B = b^2 + 2ab$

This arises from the norm-multiplicativity $\rm\ N(\alpha^2) = (N(\alpha))^2\ $ for $\alpha$ an Eisenstein integer.

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Thanks this was really useful. –  elang Dec 15 '13 at 22:56

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