Under what conditions for $a$ and $b$ are $(1)$ and $(2)$ integers? $$\sqrt{a^2-4b}\tag{1}$$ $$\sqrt{b^2-4a}\tag{2}$$ Apologies, but my only attempt is to show that $a^2-4b$ and $b^2-4a$ are perfect squares, but even on that I have no idea how to approach.
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2Did you by any chance come from here? – TheSimpliFire Mar 11 '18 at 16:40
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Unfortunately, no. But I can see why you would say that. :) – John Glenn Mar 11 '18 at 16:44
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An answer to this question would solve the one in the link :) – TheSimpliFire Mar 11 '18 at 16:46
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Yes I see that :) This tells me to use modular arithmetic, binary search or the integer version of newton's method, but I have no idea to use them. – John Glenn Mar 11 '18 at 16:50
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I have found one condition $ab\ge 16$ but i don't think it would be complete. – dssknj Mar 11 '18 at 17:28
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@Tal-Botvinnik The statement "$mn$ is a square if and only if $m$ and $n$ are squares" is not true. You need the condition that $\gcd(m,n)=1$. – ajotatxe Mar 11 '18 at 17:38
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Right, my bad. For instance 100=20*5. I'll remove my misleading comment – Tal-Botvinnik Mar 11 '18 at 17:39
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The use of modular arithmetic is very compelling, but i havent any idea how to use it. Haha – John Glenn Mar 11 '18 at 17:52
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@JohnGlenn Are $a$ and $b$ positive integers ? – Peter Mar 11 '18 at 21:35
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Yes ${}{}{}{}{}{}$ – John Glenn Mar 11 '18 at 21:39
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Numerically it seems as if $(4,4)$, $(5,6), and $(6,5)$ are the only solutions. – Carl Schildkraut Mar 11 '18 at 23:26
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Is that by trial and error? Or could there possibly exist infinitely many combinations for A and B? – John Glenn Mar 11 '18 at 23:28
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please refine question based on answers you have given in comments. – James Arathoon Mar 12 '18 at 00:14
1 Answers
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We claim the only solutions are $(4,4)$, $(5,6)$, and $(6,5)$.
Without loss of generality, let $a\leq b$. Then either $b\leq 4$ or
$$(b-3)^2 <b^2-4b \leq b^2-4a \leq b^2.$$
(we have that $(b-3)^2<b^2-4b$ if $b>9/2$.) The case where $b\leq 4$ can be hand-bashed out to reveal only the solution $\boxed{(4,4)}$. Otherwise, as $b^2-4a$ is a square between $(b-3)^2$ and $b^2$, it is either $(b-2)^2$ or $(b-1)^2$. However, we have
$$b^2-4a=(b-1)^2=b^2-2b+1 \implies 4a=2b-1,$$
which is a contradiction since $2b-1$ is odd and thus not a multiple of $4$. So, $b^2-4a=(b-2)^2$, which gives $a=b-1$. Then, we have that
$$a^2-4b=(b-1)^2-4b=b^2-6b+1$$
is a square. However, if $b\geq 15/2$, we have that
$$(b-4)^2=b^2-8b+16 < b^2-6b+1 < b^2-6b+9 = (b-3)^2,$$
a contradiction. Thus, $b\leq 15/2$, reducing the problem to a finite case check revealing only $\boxed{(5,6)}$, and correspondingly $\boxed{(6,5)}$.
Carl Schildkraut
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@JamesArathoon I haven't (per the comment), but after some testing it seems as if the only solutions are those with $a+b=-1$ and $a=-n^2,b=0$ or vice versa. – Carl Schildkraut Mar 12 '18 at 00:29
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@JamesArathoon Just an addendum; it's actually not that hard to prove that the solutions mentioned in my previous comment are the only ones where one of ${a,b}$ is allowed to be non-positive: just split into cases and use the fact that there are no positive integers $a,b$ such that $a^2+4b$ is a square and $b\leq a$. – Carl Schildkraut Mar 12 '18 at 01:55