Finding largest prime that divides an integer that is equal to sum of the squares of its four smallest positive divisors

Question

Sorry about the title... wasn't sure what to put?? My question is: A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer?

I have played around with it a bit, and have got no where. I was hoping to not need to just sub values in or solve intuitively. How do you solve this question, and is there a general route to this type of question?

Answer 1

Let $n$ be that integer and $1,u,v,w,\ldots, n$ its positive divisors in ascending order. Then we have $$\tag1 n=1+u^2+v^2+w^2.$$ Assume $n$ is odd. Then $u,v,w$ are odd and $1+u^2+v^2+w^2$ is even, contradiction. Therefore $n$ is even and hence $u=2$. Now $v^2+w^2=n-5$ is odd, hence one of $v,w$ is even and the other is odd. The following cases are possible:

• $v=4$ and $w$ is odd (and $\ge5$). Then $w^2+1\equiv 2\pmod 8$, whereas $n-u^2-v^2$ is a multiple of $4$, contradiction
• $v=3$, $w=4$. We arrive at the same contradiction
• $v$ is an odd prime and $w=2v$. Then $n=1+4+v^2+4v^2=5(1+v^2)$ is a multiple of $5$. As $w\ge 6$, we conclude $v=5$, $n=130$.

Checking the only solution $n=130$, we find that the largest prime divisor is $13$.