Sums of Two Squares

Summary: Numbers expressible as $x^2+y^2$ with integer $x,y$. Closed under multiplication via the brahmagupta-fibonacci-identity. Fermat’s two-square theorem (asserted by Euler in §172, footnote 83): an odd prime $p$ is a sum of two squares iff $p\equiv 1(\mathrm{mod}4)$. No number $\equiv 3(\mathrm{mod}4)$ is a sum of two squares.

Sources: chapter-2.0.11, chapter-2.0.12, chapter-2.0.14

Last updated: 2026-05-09

---

The Set

Numbers of the form $x^2+y^2$ up to 50 (with $\gcd (x,y)$ unrestricted, so $0$ is allowed):

$1,2,4,5,8,9,10,13,16,17,18,20,25,26,29,32,34,36,37,40,41,45,49,50.$

Euler’s classification (ch2.0.11-quadratic-form-factorization §172) splits these by structure:

• Compound: products of multiple smaller representable numbers (built via the identity below).
• Simple: irreducibly representable: $1,2,5,9,13,17,29,37,41,49,\dots$

The simple list contains:

• the prime $2$, and
• odd primes $\equiv 1(\mathrm{mod}4)$: $5,13,17,29,37,41$, and
• squares of primes $\equiv 3(\mathrm{mod}4)$: $9=3^2$, $49=7^2$.

---

Mod-4 Obstruction (§172)

Squares modulo 4 take values $0$ (even square) or $1$ (odd square). So a sum of two squares is one of $\{0,1,2\}(\mathrm{mod}4)$. No number $\equiv 3(\mathrm{mod}4)$ is a sum of two squares.

In particular, $4n-1$ (or equivalently $4n+3$) primes like $3,7,11,19,23,31,43,47,\dots$ cannot be sums of two squares.

This is a special case of the residue-class technique systematized in ch2.0.5-impossibility-quadratic-squares and quadratic-residues.

---

Multiplicative Closure

The brahmagupta-fibonacci-identity gives:

$(p^2+q^2)(r^2+s^2)=(pr-qs{)}^2+(ps+qr{)}^2=(pr+qs{)}^2+(ps-qr{)}^2.$

So if $m$ and $n$ are each sums of two squares, then so is $mn$. A product of $k$ sum-of-two-squares numbers gives up to $2^{k-1}$ representations of the product.

Example (§171): $5\cdot 13\cdot 17=1105$ has four representations:

$1105=33^2+4^2=32^2+9^2=31^2+12^2=24^2+23^2.$

---

Fermat’s Two-Square Theorem

Euler asserts in §172 (footnote 83):

Every prime number of form $4n+1$ is a sum of two squares.

Combined with the obstruction above and the multiplicative closure:

Theorem (Fermat, proved by Euler 1747). A positive integer $n$ is a sum of two squares $⟺$ in the prime factorization of $n$, every prime $\equiv 3(\mathrm{mod}4)$ appears to an even power.

Euler comments in §172: “this is undoubtedly true, but it is not easy to demonstrate it” — referencing his own published proof, which uses descent and the multiplicativity above.

---

As Higher Powers (Cubes, Biquadrates, Fifth Powers)

ch2.0.12-quadratic-form-as-power derives explicit parametrizations for when $x^2+y^2$ is an $n$-th power, by raising $(p+q\sqrt{-1})$ to power $n$ and reading parts:

Power	$x$	$y$	Example
$\square$	$p^2-q^2$	$2pq$	$3^2+4^2=5^2$
${\square }^3$	$p^3-3p{q}^2$	$3p^{2}q-q^3$	$2^2+11^2=125=5^3$
${\square }^4$	$p^4-6p^2{q}^2+q^4$	$4p^{3}q-4p{q}^3$	$7^2+24^2=625=5^4$
${\square }^5$	$p^5-10p^3{q}^2+5p{q}^4$	$5p^{4}q-10p^2{q}^3+q^5$	$38^2+41^2=3125=5^5$

---

Variants: $x^2+2y^2$ and $x^2+3y^2$

Euler studies analogous representability:

• $x^2+2y^2$: prime $p$ representable $⟺$ $p\equiv 1$ or $3(\mathrm{mod}8)$ (§174).
• $x^2+3y^2$: covered by the cube parametrization in ch2.0.12-quadratic-form-as-power §189.

These are special cases of the binary quadratic form theory developed by Lagrange and Gauss.

---

Application: When can $a\pm x$ both be squares?

ch2.0.14-questions-squares §217-218 derives a clean equivalence: $a+x$ and $a-x$ are simultaneously squares for some rational $x$ iff $a$ is a sum of two squares. Reason: if $a+x=p^2$ and $a-x=q^2$, then $2a=p^2+q^2$, so $2a$ is a sum of two squares — and the multiplicative theory shows $a$ must be too.

Thus all numbers $1\le a\le 50$ for which the question is impossible are precisely

$3,6,7,11,12,14,15,19,21,22,23,24,27,28,30,31,33,35,38,39,42,43,44,46,47,48,$

and the remaining $a$ all admit infinite parametric solutions via brahmagupta-fibonacci-identity.

Theorem: $p^2+q^2$ and $p^2+3q^2$ Never Both Squares (§229-230)

ch2.0.14-questions-squares §229-230 proves the impossibility of simultaneously making $p^2+q^2$ and $p^2+3q^2$ squares for coprime $p,q$. The proof:

1. Residue classes mod 4, 3, 5 force $p$ odd, $p$ not divisible by 3 or 5, $q$ divisible by $4\cdot 3\cdot 5=60$.
2. Setting $q=60m$, the constraint reduces to factoring $7200m^2$ into coprime $f,g$ of opposite parity with $f-g$ a sum of two squares.
3. For $m=1,2$ all candidates fail; descent extends to all $m$.

This is one of Euler’s earliest “genus” results — distinguishing forms $p^2+q^2$ (genus I) from $p^2+3q^2$ (genus II) in the binary-quadratic-form sense later systematized by Lagrange and Gauss.

Counter-class: Sums of Two Biquadrates

By stark contrast, ch2.0.13-impossibility-biquadrate-sums proves:

$x^4+y^4=z^2$

has no nontrivial solutions — even though $x^2+y^2=z^2$ has the infinite Pythagorean family. The jump from second to fourth powers introduces an obstruction defeated only by descent.

---

Related pages

• brahmagupta-fibonacci-identity
• ch2.0.11-quadratic-form-factorization
• ch2.0.12-quadratic-form-as-power
• ch2.0.13-impossibility-biquadrate-sums
• ch2.0.14-questions-squares
• ch2.0.5-impossibility-quadratic-squares
• pythagorean-triples
• quadratic-residues
• imaginary-numbers
• infinite-descent