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?

Answer 1

You know the smallest divisor is $1$. So there are $4$ possibilities: $1$ odds $3$ evens $2$ odds $2$ evens $3$ odds $1$ even $4$ odds $0$ evens $4$ odds $0$ evens: The sum of $4$ odd squares is always a multiple of $4$ (take them mod $4$) but since $2$ isn't one of the smallest divisors, this isn't possible. $3$ odds $1$ even: This isn't possible because the sum of $3$ odds and an even (take mod $4$) is always odd so an even number can't be a factor. $1$ odd $3$ evens: This isn't possible for the same reason as above. So there must be two odds and two evens. Obviously, the two smallest factors are $1$ and $2$. So the $4$ smallest factors are like this: $(1, 2, a, b)$ where $a$ is odd and $b$ is even. Because a is one of the smallest factors, $a$ must be an odd prime. The sum of two odds and two evens = $2$ (mod $4$), so $b$ must be $2$ times an odd number. Since it is the 4th smallest factor, it must be $2a$. So we now have this: $(1, 2, a, 2a)$ where $a$ is a prime. Adding together the squares, we have $5 + 5a^{2}$ or $5(a^{2} + 1)$ We now know that it is a factor of $5$. Therefore, it can't be a multiple of 3 because 2a would then be bigger than 5. (also, a^{2} can't equal $2$ (mod $3$)) So we know a equals $5$ and $2a = 10$. We have $(1,2,5,10)$. Adding together the squares, we get $130$, which has a largest prime factor of $13$.