Foxy asks whether it’s possible to have a 3*3 magic square all of whose entries are square integers. It’s fairly elementary to show that in a 3*3 magic square, the common sum is three times the middle entry. So let’s say the entries of the magic square go [*a*^2, *b*^2, *c*^2; *d*^2, *e*^2; *f*^2; *g*^2, *q*^2, *h*^2]; we want to have, for example, *a*^2 + *h*^2 = 2*e*^2.

This alone is very restrictive, since it requires (*a*^2, *e*^2, *h*^2) to be an arithmetic sequence, a sufficiently special situation that unless *a* = *e* = *h*, the next term in the sequence will never be a square (no, I can’t prove it; I’ve tried). So to find such magic squares whose entries are not all the same, it makes sense to start bashing through the Diophantine equation *a*^2 + *h*^2 = 2*e*^2.

First, the easy part: we can assume *a*, *e*, and *h* are coprime; otherwise, divide by their greatest common divisor. If *e* is even, then 2*e*^2 is divisible by 8, in which case it’s impossible for *a* and *h* to be odd since odd squares are equivalent to 1 mod 8, so that *a* and *h* are even and the three numbers are not coprime. So *e* is odd, from which it’s easy to show that so are *a* and *h*.

Now, when we have sums of two squares, the most natural environment to study divisibility is **Z**[*i*], the ring of Gaussian integers. We get (*a* + *hi*)(*a* – *hi*) = 2*e*^2. As *a* and *h* are odd, *a* (+/-) *hi* is not divisible by 2, so both *a* + *hi* and *a* – *hi* must be divisible by 1 + *i*. That condition isn’t a problem: (*a* + *hi*)/(1 + *i*) = (*a* + *hi*)(1 – *i*)/2 = (*a* + *h*)/2 + *i*(*h* – *a*)/2, and similarly (*a* – *hi*)/(1 – *i*) = (*a* + *h*)/2 + *i*(*a* – *h*)/2.

It now makes sense to write *x* = (*a* + *h*)/2, *y* = (*a* – *h*)/2. Conversely, *a* = *x* + *y* and *h* = *x* – *y*; as *a* and *h* are odd, this means that *x* and *y* have opposite parities. Since dividing once by 1 + *i* and once by 1 – *i* factors out 2 on the right-hand side, this turns the equation into (*x* + *iy*)(*x* – *iy*) = *e*^2, or *x*^2 + *y*^2 = *e*^2.

Here is where things get nightmarishly complicated. Any common factor of *x* and *y* divides *a*, *e*, and *h*, so *x*, *y*, and *e* are coprime. Consider the primes of **Z**[*i*] that divide *x* + *iy*; none can be in **Z** because that would imply a common factor of *x* and *y*. But all primes in **Z**[*i*] that have no associate in **Z** have norm equivalent to 1 or 2 mod 4, where here 1 + *i* is impossible because 1 + *i* divides *x* + *iy* iff *x* and *y* have the same parity. So right off the bat, *e* is the product of primes equivalent to 1 mod 4.

Further, each distinct prime factor after the first of *e* gives exactly two choices of *x* and *y*. To see what I mean, first consider the case when *e* is prime. Then *e* = (*r* + *is*)(*r* – *is*), so *e*^2 = (*r* + *is*)(*r* – *is*)(*r* + *is*)(*r* – *is*). There are two ways of grouping them into *x* and *y*, but the one that sends the first two to *x* + *iy* factors *e*^2 as *e***e*, i.e. *x* = *e* and *y* = 0. Hence there’s only one way, the one that sends (*r* + *is*)^2 to *x* + *iy*. If *e* is divisible by an additional prime and breaks as (*r*1 + *is*1)(*r*1 – *is*1)(*r*2 + *is*2)(*r*2 – *is*2) then once the decision to send (*r*1 + *is*1)^2 to *x* + *iy* has been made, it’s possible to send (*r*2 + *is*2)^2 or (*r*2 – *is*2)^2 to *x* + *iy*. This easily generalizes to further prime factors, but not to repeated factors.

This means that *e* can’t be prime because then the eight elements of the magic square around it will all be either *a*^2 or *h*^2, making it impossible for the other magic square conditions to hold. Now, using capital letters to denote squares, we get that the magic square is [A, B, 3E – A – B; 4E – 2A – B, E, 2A + B – 2E; A + B – E, 2E – B, 2E – A].

The above considerations show that E must be divisible only by primes equivalent to 1 mod 4 because it’s sandwiched between A and 2E – A. But by the same token, 2E – A is sandwiched between 4E – 2A – B and B, so it must similarly be divisible only by primes equivalent to 1 mod 4; similar considerations apply to A, which is sandwiched between 2A + B – 2E and 2E – B.