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Suppose $n$ is a perfect square. Consider the set of all numbers which are the product of two numbers, not necessarily distinct, both of which are at least $n$. Express the $n$th smallest number in this set in terms of $n$.

If, for example, $n = 4$, then the numbers are $4^2, 4 \cdot 5, 4 \cdot 6, 5^2$. If $n = 9$, then the numbers are $9^2,9 \cdot 10, 9 \cdot 11, 10^2, 9 \cdot 12, 10 \cdot 11,9 \cdot 13, 10 \cdot 12, 11^2$. Thus we conjecture the answer is $(n+\sqrt{n}-1)^2$, but how do we prove this?

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  • $\begingroup$ that number isn't an integer for most $n$ I think. $\endgroup$ – Jorge Fernández Hidalgo Jan 18 '17 at 3:39
  • $\begingroup$ @JorgeFernándezHidalgo Recall that $n$ is a perfect square. $\endgroup$ – Puzzled417 Jan 18 '17 at 3:39
  • $\begingroup$ oh, my bad.${}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Jan 18 '17 at 3:39
  • $\begingroup$ Why do you think that number works? $\endgroup$ – Jorge Fernández Hidalgo Jan 18 '17 at 3:46
  • $\begingroup$ @JorgeFernándezHidalgo Because I tested it for some examples and it was true. $\endgroup$ – Puzzled417 Jan 18 '17 at 3:48
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Clearly there $w=(n+\sqrt n -1)^2$ is at least the $n$'th number in the list.

We must only prove that $w<ab$ if $a+b>2n+2\sqrt{n}-2$, with both $a$ and $b\geq n$.

Clearly this product is minimized in the case $a=n$, Which leaves $b>n+2\sqrt{n}-2$, since $b$ is an integer we can take the minimal case $b=n+\sqrt{2n}-1$.

So we must only prove $n(n+2\sqrt{n}-1)\geq (n+\sqrt{n}-1)^2\iff -n>-2n-2\sqrt{n}+n+1=-n-2\sqrt{n}+1$.

So it is in fact true.

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  • $\begingroup$ Why is $w$ at least the $n$th number in the list? $\endgroup$ – Puzzled417 Jan 18 '17 at 15:35

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