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The natural numbers $a$ and $b$ are such $a^2+ab+1$ is divisible by $b^2+ba+1$. Prove that $a = b$.

I tried to algebraically manipulate it as follows:

$(b^2 + ba + 1)k = a^2 + ab + 1$

$[b(a + b) + 1]k = a(a + b) + 1$

$kb(a + b) + k = a(a + b) + 1$

$k - 1 = (a - kb)(a + b)$

I'm stuck here. What should I do next? A case-by-case analysis of possible congruencies would be too tedious and inelegant.

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up vote 3 down vote accepted

If $a^2+ab+1$ is divisible by $b^2+ba+1$, then so is $(a^2+ab+1)-(b^2+ba+1)=a^2-b^2$.

Note that $a+b$ and $b^2+ba+1$ are relatively prime. So $b^2+ab+1$ divides $a-b$. Now you should be able to finish, using considerations of size.

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we can directly write: if $d|(a^2+ab+1),(b^2+ba+1)$ $d$ must divide $a(b^2+ba+1)-b(a^2+ab+1)=a-b$ – lab bhattacharjee Jun 26 '13 at 3:38
@labbhattacharjee: Yes, I got there a slower way. – André Nicolas Jun 26 '13 at 3:39
Thanks! Nice solution. – Gerard Jun 26 '13 at 3:40
@AndréNicolas: hi, i just have a question, can you help me a little bit more with how you continue from where you left off? $b^2+ab+1$ divides a-b – Akaichan Oct 1 '13 at 19:22
We have $0\le a-b$. Note that $ab\ge a\gt a-b$. So $b^2+ab+1\gt a-b$. The only way a "big" number like $b^2+ab+1$ can divide a number like $a-b$ which has smaller absolute value is if $a-b=0$. – André Nicolas Oct 1 '13 at 19:36

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