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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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    that number isn't an integer for most $n$ I think. – 2017-01-18
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    @JorgeFernándezHidalgo Recall that $n$ is a perfect square. – 2017-01-18
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    oh, my bad.${}{}{}$ – 2017-01-18
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    Why do you think that number works? – 2017-01-18
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    @JorgeFernándezHidalgo Because I tested it for some examples and it was true. – 2017-01-18
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    Do you know the GM-AM. (n+k)(n+k) < (n+(k-a))(n+(k+a)) the numbers in order will be n, (n+1)n,(n+1)(n+1),n (n+2), (n+1)(n+2),n (n+3) and so on. Can you take it from there. – 2017-01-18

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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 $w2n+2\sqrt{n}-2$, 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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    Why is $w$ at least the $n$th number in the list? – 2017-01-18