Problem:
Let $a,b\in\Bbb N$ with $a\cdot(a,b)=b\cdot[a,b]$ where (a,b) means the greatest common divisor of a,b and [a,b] means the smallest common multiple of a,b prove that there are an infinite amount of solutions for a,b.
What I tried:
I first started off by assuming WLOG that $a<b$ and then assuming 1 case,that is $(a,b)=a$ and $[a,b]=b$ with that in mind we get $a^2=b^2$ and that clearly has infinite solutions,but i don't know if this is a legitimate proof and i would like to find out what's the correct proof for this if it's not this one.
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sai-kartik
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Okarine
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2Whenever we have $b\mid a$, the given equation holds. – Peter May 08 '20 at 11:32
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If $1\leq a<b$, then $a^2<b^2$. – Wuestenfux May 08 '20 at 11:38
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This is not an informative title. – May 08 '20 at 12:07
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Because of $(a,b)[a,b]=ab$, see Easiest and most complex proof of $\gcd (a,b) \times \operatorname{lcm} (a,b) =ab.$, we obtain by multiplying your equation with $(a,b)$ that $$ (a,b)^2=b^2. $$ This has infinitely many solutions in $a$ and $b$ (namely all $a,b$ such that $b\mid a$).
Dietrich Burde
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