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I first tried with brute force with $-1000 \leq a,b \leq 1000$ but found no solution. But then a simple argument showed me that there was no solution. Not only in the integers, but even for the rationals. What was that argument?

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And why is it impossible to solve the equation? $a^2+ab-b^2=0$

$$a=\frac{-b\pm\sqrt{b^2+4b^2}}{2}=b\frac{-1\pm\sqrt{5}}{2}$$

you can see the whole decision no.

If $a^3+ab-b^2=0$ would have acted similarly. $b^2-ab-a^3=0$

$$b=\frac{a\pm\sqrt{a^2+4a^3}}{2}=a\frac{1\pm\sqrt{4a+1}}{2}=a\frac{1\pm{k}}{2}$$

$a$ - to choose such that the root was intact. $a=\frac{k^2-1}{4}$

If $a^2+pab-b^2=0$

$$a=\frac{-pb\pm\sqrt{p^2b^2+4b^2}}{2}=b\frac{-p\pm\sqrt{p^2+4}}{2}$$

There are solutions when: $p=0 $

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  • $\begingroup$ +1:This answer is also correct though it makes no use of arithmetics. $\endgroup$ – Marc Bogaerts Sep 5 '14 at 14:12
  • $\begingroup$ But what would you do if the equation were $a^3+ab-b^2$ or even $a^2+pab-b^2$ with p a prime? $\endgroup$ – Marc Bogaerts Sep 5 '14 at 14:15
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For rational numbers, simply write $a$ and $b$ with common denominators, and then clear denominators to get an integer solution. So if there is no integer solution, there is no rational solution.

For integers:

$$a^2+ab-b^2=0$$ $$4a^2+4ab-4b^2=0$$ $$4a^2+4ab+b^2=5b^2$$ $$(2a+b)^2=5b^2$$

Left side is a perfect square; right side is not (by Fundamental Theorem of Arithmetic). Contradiction.

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Without loss of generality we can assume that $\gcd(a,b)=1$. If we take the equation modulo $a$ (or b) we get $b^2 \equiv 0 \pmod a$, a contradiction.

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    $\begingroup$ That's not a contradiction immediately. It just forces $a = \pm 1.$ $\endgroup$ – Geoff Robinson Sep 6 '14 at 1:06
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We will prove this by checking out different cases:

Case 1: Both are odd. In that case $a^2 - b^2$ is even and $ab$ is odd. You cannot get zero then.

Case 2: $a$ is odd and $b$ is even: $a^2 - b^2$ is then odd and $ab$ is even. Again, not possible to get zero

Case 3: Same as case 2 with $b$ being odd and $a$ even.

Case 4: Let $a=2n$ and $b=2m$ then we have $a^2+ab-b^2=0 \Rightarrow 4n^2 +4nm -4m^2=0 \Rightarrow n^2+nm-m^2=0$ and if they're even then you repeat this till you obtain one of the other three cases which means none of the cases are possible.

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  • $\begingroup$ This is correct but there is a much shorter way to arrive at the result. It uses modular arithmetic. $\endgroup$ – Marc Bogaerts Sep 5 '14 at 14:06
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    $\begingroup$ You can expand this to work in the rationals. Let $a=a_n/a_d$ and $b=b_n/b_d$. It can be shown that $a^2+ab-b^2=0\iff(a_nb_d)^2+(a_nb_d)(b_na_d)-(b_na_d)^2=0$ and since $(a_nb_d), (b_na_d)$ are both integers, it's proven by the above. $\endgroup$ – BeaumontTaz Sep 5 '14 at 14:08
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    $\begingroup$ @Nimda, looking at the parity like Sheheryar did is the same as looking at $a,b\pmod{2}$. It's just a very simple version of modular arithmetic. $\endgroup$ – BeaumontTaz Sep 5 '14 at 14:13
  • $\begingroup$ It was not $\mod 2$ what I had in mind. $\endgroup$ – Marc Bogaerts Sep 5 '14 at 14:16
  • $\begingroup$ @Nimda, then can you specify in your question what it was you had in mind? Give as much as you can remember about it. $\endgroup$ – BeaumontTaz Sep 5 '14 at 14:17
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For integers it is easy, if a prime $p$ divides $a$ then it divides $b$ and conversely, so $\frac{a}{b}$ can never be a reduced fraction. For rationals multiply by the denominators to reduce to the integer case.

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  • $\begingroup$ To fill in some minor details about why $p|a\implies p|b$. Assuming $p|a$ we know that $a=0\pmod{p}$ and $a^2=0\pmod{p}$ Thus $a^2+b(a-b)=0\pmod{p}$ reduces to $b(a-b)=0\pmod{p}$ which implies either $b=0\pmod{p}$ or $a-b=0\pmod{p}$. The latter is the same as the former. So $p|b$. $\endgroup$ – BeaumontTaz Sep 5 '14 at 14:26
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    $\begingroup$ better to say $p|a$ and $a(a+b)+b^2=0$ so $p|b^2$ so $p|b$ $\endgroup$ – Rene Schipperus Sep 5 '14 at 14:28
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If there were a rational solution, we could multiply it through by the square of a suitable non-zero integer to get an integer solution, so we may suppose that $a,b \in \mathbb{Z}.$ But now if $c = {\rm gcd}(a,b)$ we can divided through by $c^{2}.$ Hence we may suppose that $a,b \in \mathbb{Z}$ and that ${\rm gcd}(a,b) = 1.$ Now, however, $b^{2} = a(a+b),$ so that $a$ divides $b^{2}$. Hence $a$ divides ${\rm gcd}(a^{2},b^{2}) = 1.$ Thus $a = \pm 1.$ Similarly, $b$ divides $a^{2}$ and $b = \pm 1.$ These two facts together yield a contradiction, as $a^{2} +ab$ is then even and $b^{2}$ is odd.

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