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How to show that every integer greater than $23$ is the sum of two squareful numbers? I checked up to $50000$. The argument I used in this answer to a similar problem doesn't work because it would rely on most numbers being squareful which isn't the case.

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What's a squareful number? – Michael Albanese Sep 3 '13 at 1:35
A squareful number is divisible by a square greater than $1$. – Dan Brumleve Sep 3 '13 at 1:36
up vote 13 down vote accepted

Every large enough $n$ can be represented as $n=4x+9y$, where $x$ and $y$ are non-negative integers. Here large enough means $\ge (4-1)(9-1)$.

For completeness, we give a proof. The four numbers $24$, $25$, $26$, and $27$ are so representable. And every integer $\ge 28$ is of the shape $4k+t$, where $t$ is one of these four numbers, and $k$ is a positive integer.

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That's really the answer, not a hint :P +1 – Doorknob Sep 3 '13 at 1:37
It could be a hint because there is no proof given... :) +1 – apnorton Sep 3 '13 at 2:03
Wait, for $n=24$, $24-9=15$, $24-2*9=6$, and $24-3*9$. So $24$ is not representable as $4x+9y$ for positive $x$ and $y$. – nayrb Sep 3 '13 at 2:43
@nayrb: Technically, you just need that $24,\dots,27$ are of the form $4k+s$, where $s$ is a squareful number, then the rest follows by induction as André described. $24=4+20$ works just fine. – Tomas Sep 3 '13 at 2:46

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