Which whole numbers can be written as a^2 + b^2 for two whole numbers a, b? Try a handful by hand: 2 = 1^2 + 1^2 works, 5 = 1^2 + 2^2 works, 13 = 2^2 + 3^2 works — but 3 refuses, and so do 7, 11 and 19, no matter how you juggle the squares. It looks like a coin toss until you line the primes up and check the remainder each one leaves when divided by 4 — and then a perfectly sharp pattern snaps into focus.

Pierre de Fermat spotted that pattern in the 1640s and, in his usual style, announced it in a letter with a proof he never bothered to write down. It took nearly a century for anyone else to nail it down properly — and when they did, the reason turned out to run much deeper than a curiosity about squares: it is one of the first signs that whole numbers and complex numbers are secretly talking to each other.

The prime case

So 5, 13, 17, 29, 37, 41 all split into two squares, while 3, 7, 11, 19, 23, 31 never do — however hard you search. The "only if" half is genuinely easy: square any whole number and reduce it modulo 4, and you only ever get 0 or 1 (an even number squared is \equiv 0, an odd number squared is \equiv 1). Add two such remainders and the only possibilities are 0, 1, or 2 \pmod 4 — 3 \pmod 4 is simply never on the menu.

The "if" direction — that every prime \equiv 1 \pmod 4 really is achievable — is the deep part Fermat claimed but never showed. It leans on the fact that -1 is a quadratic residue precisely when p \equiv 1 \pmod 4, which lets you engineer a multiple of p that is itself a sum of two squares, then shrink it down step by step until the multiple disappears.

Worked example: checking primes directly

Before trusting a theorem, test it by hand. Take 13: divide by 4 and the remainder is 1, so the theorem promises a representation. Sure enough,

And 29: 29 = 4 \cdot 7 + 1, so 29 \equiv 1 \pmod 4 — and indeed 29 = 2^2 + 5^2 = 4 + 25.

Now try the other side of the fence. Take 7: it is \equiv 3 \pmod 4, so the theorem says no pair of squares can ever add up to it. Check every possibility up to 7 — 0+7,\ 1+6,\ 4+3 — and none of those second numbers is itself a perfect square. The same happens for 11 \equiv 3 \pmod 4: the candidate splits are 1+10,\ 4+7,\ 9+2, and again nothing works. The theorem didn't just predict failure — it explained why it was inevitable, before you ever picked up a pencil.

The full classification

Multiplicativity extends the prime case to every whole number. The key is Brahmagupta's identity, discovered a thousand years before Fermat, which shows sums of two squares are closed under multiplication:

Multiply two sum-of-two-squares numbers together with this identity and the answer is automatically another sum of two squares — no searching required, the new a, b just fall out of the formula. Chase that idea through every prime factor and you land on the complete rule:

Worked example: testing composite numbers

Take 45. Factorise it first:

The only prime \equiv 3 \pmod 4 present is 3, and it appears to the power 2 — even. The rule says 45 must be a sum of two squares, and indeed

Now try 21 = 3 \cdot 7. Both prime factors, 3 and 7, are \equiv 3 \pmod 4, and each appears to the power 1 — odd. The rule predicts failure, and a quick check of every split up to 21 confirms it: no pair of squares sums to 21.

One more, to see the rule handle a busier number. Take 90 = 2 \cdot 3^2 \cdot 5. The prime 2 is unrestricted, 5 \equiv 1 \pmod 4 is unrestricted, and the only policed prime, 3, again appears squared. Sum of two squares it is:

One way, or many?

A prime p \equiv 1 \pmod 4 has, in a precise sense, only one way to split into two squares — swap a and b, or flip a sign, and you're still describing the same pair. Check 13: the only positive solution is 2^2+3^2; no other pair of positive whole numbers works.

Composite numbers can be more generous. Take 65 = 5 \cdot 13, a product of two different primes \equiv 1 \pmod 4. It has two genuinely different representations:

Each extra distinct prime factor \equiv 1 \pmod 4 you multiply in tends to unlock another representation — a direct echo of the Gaussian-integer splitting: every prime that splits contributes a fresh way to pair up its complex factors.

Counting the representations exactly

There's even an exact formula for how many ways, due to Jacobi. Let d_1(n) and d_3(n) count the divisors of n that are \equiv 1 and \equiv 3 \pmod 4 respectively.

Try it on 65: its divisors are 1, 5, 13, 65, and every single one is \equiv 1 \pmod 4, so d_1 = 4 and d_3 = 0, giving r_2(65) = 16. That matches perfectly: the two unordered positive pairs (1,8) and (4,7) each generate 8 signed, ordered variants (2 orderings times 2\times2 sign choices), for 8+8=16 in total.

A bonus: free Pythagorean triples

Every prime p = a^2+b^2 hands you a Pythagorean triple for free, no searching required. Since

the triple (a^2-b^2,\ 2ab,\ a^2+b^2) always has legs and a hypotenuse that are whole numbers, with the prime itself sitting as the hypotenuse. Feed in 5 = 1^2+2^2 (so a=2, b=1) and out drops the smallest Pythagorean triple of all:

Try 13 = 3^2+2^2: it produces (3^2-2^2,\ 2\cdot3\cdot2,\ 13) = (5, 12, 13). Every prime \equiv 1 \pmod 4 generates its own triple this way, all from the same handful of algebra that Brahmagupta used to multiply sums of squares together.