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.

Mr. Levy, you are a beast. A beast of Math.

How did you come to be like this? As it stands right now, I’m still plodding through my freshman requirements, Analysis and Intro to Number Theory. What would you recommend beyond that? Any particular textbooks you might recommend?

How do I get your kinds of superpowers?

– Foxy

Ew. I didn’t learn algebraic number theory from a single book; each book had different strengths and weaknesses, so whenever I hit a weak point, I’d move to another book. The book that I think best explains the technique of factoring in a number field is Ian Stewart and David Tall’s Algebraic Number Theory. There’s also a small part right at the beginning of Jürgen Neukirch’s book by the same title that proves that every prime p = 1 mod 4 can be written as the sum of two squares, but the rest of the book is heavily theoretical (as in, better grad students than me are stumped by it).

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