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Given the following equation: $$a\times b=y^2$$ Where a,b and y are integers. One of these two things must be true. Either both a and b are prefect squares or a and b are identical.

i) For example 2×2=4. We know 2 is not a perfect square but 2 multiplied by another 2 is a perfect square.
ii) 4×9=36. Both 4 and 9 are perfect squares and their product is a perfect square.

The question is whether these two scenarios are the only possible scenarios for a and b or is there a third scenario that I am overlooking?

I am not a number theorist and that is why I am asking whether there is any other possiblity except these two.

Thomas Andrews
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Tony Kuria
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    $2\times 8=4^2$ –  Jan 21 '22 at 06:52
  • There are other possibilities. For example $a=2$, $b=8$, $y=4$. – M. Wind Jan 21 '22 at 06:52
  • There is something similar: if you divide $a$ and $b$ by their greatest common divisor, you will get perfect squares. In other words, if $d=\gcd(a,b)$ then $a/d$, $b/d$ are squares. –  Jan 21 '22 at 06:56
  • To generalize the example given in the previous comments : if $y$ has at least two prime divisors, then there exists $a,b$ such that $ab=y^2$, where $a\neq b$ and $a$ and $b$ are not perfect squares. – TheSilverDoe Jan 21 '22 at 06:59
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    The general condition is, there exist $r,s,t$ such that $a=rs^2$ and $b=rt^2$. – Gerry Myerson Jan 21 '22 at 07:00
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    Guyz some of your comments are very helpful and should be posted as answers. Having read your comments, I gather that a and b must be coprime for the two conditions in my question to hold. Looking at 2 and 8, they share a common factor of 2. This is interesting, I have learnt something there. – Tony Kuria Jan 21 '22 at 07:07

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Your conditions are only true when $a$ and $b$ are coprime, as other possibilities like $a=2,b=8$ violates them.

Credits goes to @StinkingBishop @M.Wind @TheSilverDoe @GerryMyerson

Aaa Lol_dude
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