Prove that if integers $a,b$ satisfy the equation $2a^2+a=3b^2+b$ then numbers $a-b$ and $2a+2b+1$ are perfect squares. $a,b$ are integers . This is my proof is it correct? I'm referring to one answer from this thread $\sqrt a$ is either an integer or an irrational number. .
Proof for $a-b$.
$$a-b=3b^2-2a^2$$
Referred here: If $a^{1/b}=\frac{x}{y}$
where $y$
does not divide $x$
, then $a=(a^{1/b})^b=x^b/y^b$
is not an integer (since $y^b$
does not divide $x^b$
), giving a contradiction.
Lemma: If $y$
does not divide $x
$
, then $y^b$
does not divide $x^b$. There is also a proof of this lemma on the linked question.
It is straightforward to see that $a-b=3b^2-2a^2$ is an integer.
I don't have a proof yet for the second part.
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What if $a^{1/b}$ is an irrational number? – pi66 Aug 23 '16 at 08:44
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Then it is of the form $\frac{x}{y}$ . – I I Aug 23 '16 at 08:46
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No, I said it could be an irrational number. – pi66 Aug 23 '16 at 08:52
2 Answers
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I don't think your proof works, because $a^{1/b}$ could be an irrational number, in which case it cannot be written as $\frac{x}{y}$ with $x,y$ being integers. Moreover, I'm not sure why you would be interested in the quantity $a^{1/b}$.
For a hint, note that $$(a-b)(2a+2b+1)=2a^2-2b^2+a-b=b^2.$$ What is the greatest common divisor of $a-b$ and $2a+2b+1$? What can you conclude?
pi66
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Assume that a prime number $p$ divides both $a-b$ and $2a+2b+1$. What can you get? – pi66 Aug 23 '16 at 09:41
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Suppose $p$ divides $a-b$. From the equation $(a-b)(2a+2b+1)=b^2$, what other term must it divide? – pi66 Aug 23 '16 at 10:00
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Your proof is not correct for simple reasons: If it were correct, it would prove too much. To see this, note that it seems that your argument woiuld work just as fine if the equation were $$ 2a^2+a = 11b^2+b$$ instead. But this equation holds for $a=3$, $b=1$ and $a-b=2$ is not a square.
Hagen von Eitzen
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