I recently happened to notice that if
(a^2) + (b^2) = 2p,where a, b, and p are all integers, then
((a+b)/2)^2 + ((a-b)/2)^2 = p.
For example, if we know that
5^2 + 1^2 = 26 = 2*13,
it follows that
((5+1)/2)^ + (5-1)/2)^2 = 13,
that is,
3^2 + 2^2 = 13.
Suppose, though, that
(a^2) + (b^2) = 3p
or, more generally,
(a^2) + (b^2) = op,
where o is an odd integer.
Is there some way of deriving from this a formula such that
(c^2) + (d^2) = p,
where c and d are also integers, with values determined by a and b?
You rang? Here's a quick look.
ReplyDeleteLet's examine the case for a=3 and b=9.
a^2 + b^2 = 9 + 81 = 90.
The odd divisors of 90 are 1, 3, 5, 9, 15 and 45.
Dividing these into 90 give p-values of 90, 30, 16, 6, and 2. Apart from the trivial case of o=1, none of the others are sums of squares of integers. Does this particular case invalidate the search for a general formula of some kind? I'd have to have a closer look tomorrow I think.
Thanks!
ReplyDeleteMaybe you're right that there's no general formula. But if p is prime and p-1 divisible by 4, we know that p is the sum of two squares. And it's also true that there must be a pair of numbers ±a that are square roots of -1 mod p, where 0 < a < p/2. (Does everybody know that? I stumbled on it a few weeks ago, and I think I can prove it.) If we know the two squares whose sum is p, we can find the square roots of -1 easily enough.
Suppose that
a^2 + b^2 = p.
Shifting to mod p, we have
a^2 + b^2 = 0 {allow me to substitute equal signs for congruence signs}, whence
a^2 = -(b^2), and
(a/b)^2 = (b/a)^2 = -1.
For example,
5^2 + 2^2 = 29,
so 5/2 and 2/5, which come out to -12 and 12, respectively, are the square roots of -1 mod 29.
Knowing that, we can be sure that
12^2 + 1^2 = -1 + 1 = 0 (mod 29).
By way of confirmation,
12^2 + 1^2 = 145 = 5*29.
But can we go the other way and find the pair 5, 2 from the pair 12, 1? Maybe not . . .