Real square roots of –1

Real Square Roots of –1

Yes, well I quite like imaginary and complex numbers, starting with i and –i, the square roots of –1.
But did you know that –1 also has square roots that are real numbers?
Well it does.
Sometimes.
In modular arithmetic.

In particular, if p is prime and congruent to 1 modulo 4, then –1 has a pair of real integer square roots modulo p.

Why? 

Consider that there are p – 1 distinct non-zero numbers in modulus p. 

If p is prime, each has a unique reciprocal (allow me to spare you the proof, which is simple but I’d probably mess up). 

And the reciprocal of the inverse of each is the inverse of the reciprocal. 

So we have sets of four numbers, a, –a, b, –b, such that 1/a = b and 1/(–a) = –b. 

But if a = 1, b =1, and of course –a = –1 and b = –1. 

Now, if p ≡ 1 (mod 4), p – 1 is divisible by 4, meaning that 1 and –1 must be part of a foursome with another pair of numbers. 

And since these numbers cannot be their own reciprocals (only 1 and –1 can manage that), they must be reciprocal to each other. 

IOW, a = 1/(–a) and –a = 1/a. 

Whence a²= ­–1 and (–a)² = –1.

So there you have it. 

Playing with sums of two squares

Yes, well . . .

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?