Fermat's theorem on sums of two squares

Given an odd prime ${\displaystyle p}$ in the form ${\displaystyle 4k+1}$, first find ${\displaystyle x}$ such that ${\displaystyle x^{2}\equiv -1{\pmod {p}}}$. This can be done by finding a Quadratic non-residue modulo ${\displaystyle p}$, say ${\displaystyle q}$, and letting ${\displaystyle x=q^{\frac {p-1}{4}}{\pmod {p}}}$.

Such an ${\displaystyle x}$ will satisfy the condition since quadratic non-residues satisfy ${\displaystyle q^{\frac {p-1}{2}}\equiv -1{\pmod {p}}}$.

Once ${\displaystyle x}$ is determined, one can apply the Euclidean algorithm with ${\displaystyle p}$ and ${\displaystyle x}$. Denote the first two remainders that are less than the square root of ${\displaystyle p}$ as ${\displaystyle a}$ and ${\displaystyle b}$. Then it will be the case that ${\displaystyle a^{2}+b^{2}=p}$.

Take ${\displaystyle p=97}$. A possible quadratic non-residue for 97 is 13, since ${\displaystyle 13^{\frac {97-1}{2}}\equiv -1{\pmod {97}}}$. so we let ${\displaystyle x=13^{\frac {97-1}{4}}=22{\pmod {97}}}$. The Euclidean algorithm applied to 97 and 22 yields:

$${\displaystyle 97=22(4)+9,}$$

$${\displaystyle 22=9(2)+4,}$$

$${\displaystyle 9=4(2)+1,}$$

$${\displaystyle 4=1(4).}$$

The first two remainders smaller than the square root of 97 are 9 and 4; and indeed we have ${\displaystyle 97=9^{2}+4^{2}}$, as expected.

Euler succeeded in proving Fermat's theorem on sums of two squares in 1749, when he was forty-two years old. He communicated this in a letter to Goldbach dated 12 April 1749.[10] The proof relies on infinite descent, and is only briefly sketched in the letter. The full proof consists in five steps and is published in two papers. The first four steps are Propositions 1 to 4 of the first paper[11] and do not correspond exactly to the four steps below. The fifth step below is from the second paper.[12][13]

For the avoidance of ambiguity, zero will always be a valid possible constituent of "sums of two squares", so for example every square of an integer is trivially expressible as the sum of two squares by setting one of them to be zero.

1. The product of two numbers, each of which is a sum of two squares, is itself a sum of two squares.

This is a well-known property, based on the identity

${\displaystyle (a^{2}+b^{2})(p^{2}+q^{2})=(ap+bq)^{2}+(aq-bp)^{2}}$

due to Diophantus.

2. If a number which is a sum of two squares is divisible by a prime which is a sum of two squares, then the quotient is a sum of two squares. (This is Euler's first Proposition).

Indeed, suppose for example that ${\displaystyle a^{2}+b^{2}}$ is divisible by ${\displaystyle p^{2}+q^{2}}$ and that this latter is a prime. Then ${\displaystyle p^{2}+q^{2}}$ divides

${\displaystyle (pb-aq)(pb+aq)=p^{2}b^{2}-a^{2}q^{2}=p^{2}(a^{2}+b^{2})-a^{2}(p^{2}+q^{2}).}$

Since ${\displaystyle p^{2}+q^{2}}$ is a prime, it divides one of the two factors. Suppose that it divides ${\displaystyle pb-aq}$. Since

${\displaystyle (a^{2}+b^{2})(p^{2}+q^{2})=(ap+bq)^{2}+(aq-bp)^{2}}$

(Diophantus's identity) it follows that ${\displaystyle p^{2}+q^{2}}$ must divide ${\displaystyle (ap+bq)^{2}}$. So the equation can be divided by the square of ${\displaystyle p^{2}+q^{2}}$. Dividing the expression by ${\displaystyle (p^{2}+q^{2})^{2}}$ yields:

${\displaystyle {\frac {a^{2}+b^{2}}{p^{2}+q^{2}}}=\left({\frac {ap+bq}{p^{2}+q^{2}}}\right)^{2}+\left({\frac {aq-bp}{p^{2}+q^{2}}}\right)^{2}}$

and thus expresses the quotient as a sum of two squares, as claimed.

On the other hand if ${\displaystyle p^{2}+q^{2}}$ divides ${\displaystyle pb+aq}$, a similar argument holds by using the following variant of Diophantus's identity:

${\displaystyle (a^{2}+b^{2})(q^{2}+p^{2})=(aq+bp)^{2}+(ap-bq)^{2}.}$

3. If a number which can be written as a sum of two squares is divisible by a number which is not a sum of two squares, then the quotient has a factor which is not a sum of two squares. (This is Euler's second Proposition).

Suppose ${\displaystyle q}$ is a number not expressible as a sum of two squares, which divides ${\displaystyle a^{2}+b^{2}}$. Write the quotient, factored into its (possibly repeated) prime factors, as ${\displaystyle p_{1}p_{2}\cdots p_{n}}$ so that ${\displaystyle a^{2}+b^{2}=qp_{1}p_{2}\cdots p_{n}}$. If all factors ${\displaystyle p_{i}}$ can be written as sums of two squares, then we can divide ${\displaystyle a^{2}+b^{2}}$ successively by ${\displaystyle p_{1}}$, ${\displaystyle p_{2}}$, etc., and applying step (2.) above we deduce that each successive, smaller, quotient is a sum of two squares. If we get all the way down to ${\displaystyle q}$ then ${\displaystyle q}$ itself would have to be equal to the sum of two squares, which is a contradiction. So at least one of the primes ${\displaystyle p_{i}}$ is not the sum of two squares.

4. If ${\displaystyle a}$ and ${\displaystyle b}$ are relatively prime positive integers then every factor of ${\displaystyle a^{2}+b^{2}}$ is a sum of two squares. (This is the step that uses step (3.) to produce an 'infinite descent' and was Euler's Proposition 4. The proof sketched below also includes the proof of his Proposition 3).

Let ${\displaystyle a,b}$ be relatively prime positive integers: without loss of generality ${\displaystyle a^{2}+b^{2}}$ is not itself prime, otherwise there is nothing to prove. Let ${\displaystyle q}$ therefore be a proper factor of ${\displaystyle a^{2}+b^{2}}$, not necessarily prime: we wish to show that ${\displaystyle q}$ is a sum of two squares. Again, we lose nothing by assuming ${\displaystyle q>2}$ since the case ${\displaystyle q=2=1^{2}+1^{2}}$ is obvious.

Let ${\displaystyle m,n}$ be non-negative integers such that ${\displaystyle mq,nq}$ are the closest multiples of ${\displaystyle q}$ (in absolute value) to ${\displaystyle a,b}$ respectively. Notice that the differences ${\displaystyle c=a-mq}$ and ${\displaystyle d=b-nq}$ are integers of absolute value strictly less than ${\displaystyle q/2}$: indeed, when ${\displaystyle q>2}$ is even, gcd${\displaystyle (a,q/2)=1}$; otherwise since gcd${\displaystyle (a,q/2)\mid q/2\mid q\mid a^{2}+b^{2}}$, we would also have gcd${\displaystyle (a,q/2)\mid b}$.

Multiplying out we obtain

${\displaystyle a^{2}+b^{2}=m^{2}q^{2}+2mqc+c^{2}+n^{2}q^{2}+2nqd+d^{2}=Aq+(c^{2}+d^{2})}$

uniquely defining a non-negative integer ${\displaystyle A}$. Since ${\displaystyle q}$ divides both ends of this equation sequence it follows that ${\displaystyle c^{2}+d^{2}}$ must also be divisible by ${\displaystyle q}$: say ${\displaystyle c^{2}+d^{2}=qr}$. Let ${\displaystyle g}$ be the gcd of ${\displaystyle c}$ and ${\displaystyle d}$ which by the co-primeness of ${\displaystyle a,b}$ is relatively prime to ${\displaystyle q}$. Thus ${\displaystyle g^{2}}$ divides ${\displaystyle r}$, so writing ${\displaystyle e=c/g}$, ${\displaystyle f=d/g}$ and ${\displaystyle s=r/g^{2}}$, we obtain the expression ${\displaystyle e^{2}+f^{2}=qs}$ for relatively prime ${\displaystyle e}$ and ${\displaystyle f}$, and with ${\displaystyle s<q/2}$, since

${\displaystyle qs=e^{2}+f^{2}\leq c^{2}+d^{2}<\left({\frac {q}{2}}\right)^{2}+\left({\frac {q}{2}}\right)^{2}=q^{2}/2.}$

Now finally, the descent step: if ${\displaystyle q}$ is not the sum of two squares, then by step (3.) there must be a factor ${\displaystyle q_{1}}$ say of ${\displaystyle s}$ which is not the sum of two squares. But ${\displaystyle q_{1}\leq s<q/2<q}$ and so repeating these steps (initially with ${\displaystyle e,f;q_{1}}$ in place of ${\displaystyle a,b;q}$, and so on ad infinitum) we shall be able to find a strictly decreasing infinite sequence ${\displaystyle q,q_{1},q_{2},\ldots }$ of positive integers which are not themselves the sums of two squares but which divide into a sum of two relatively prime squares. Since such an infinite descent is impossible, we conclude that ${\displaystyle q}$ must be expressible as a sum of two squares, as claimed.

5. Every prime of the form ${\displaystyle 4n+1}$ is a sum of two squares. (This is the main result of Euler's second paper).

If ${\displaystyle p=4n+1}$, then by Fermat's Little Theorem each of the numbers ${\displaystyle 1,2^{4n},3^{4n},\dots ,(4n)^{4n}}$ is congruent to one modulo ${\displaystyle p}$. The differences ${\displaystyle 2^{4n}-1,3^{4n}-2^{4n},\dots ,(4n)^{4n}-(4n-1)^{4n}}$ are therefore all divisible by ${\displaystyle p}$. Each of these differences can be factored as

${\displaystyle a^{4n}-b^{4n}=\left(a^{2n}+b^{2n}\right)\left(a^{2n}-b^{2n}\right).}$

Since ${\displaystyle p}$ is prime, it must divide one of the two factors. If in any of the ${\displaystyle 4n-1}$ cases it divides the first factor, then by the previous step we conclude that ${\displaystyle p}$ is itself a sum of two squares (since ${\displaystyle a}$ and ${\displaystyle b}$ differ by ${\displaystyle 1}$, they are relatively prime). So it is enough to show that ${\displaystyle p}$ cannot always divide the second factor. If it divides all ${\displaystyle 4n-1}$ differences ${\displaystyle 2^{2n}-1,3^{2n}-2^{2n},\dots ,(4n)^{2n}-(4n-1)^{2n}}$, then it would divide all ${\displaystyle 4n-2}$ differences of successive terms, all ${\displaystyle 4n-3}$ differences of the differences, and so forth. Since the ${\displaystyle k}$th differences of the sequence ${\displaystyle 1^{k},2^{k},3^{k},\dots }$ are all equal to ${\displaystyle k!}$ (Finite difference), the ${\displaystyle 2n}$th differences would all be constant and equal to ${\displaystyle (2n)!}$, which is certainly not divisible by ${\displaystyle p}$. Therefore, ${\displaystyle p}$ cannot divide all the second factors which proves that ${\displaystyle p}$ is indeed the sum of two squares.

Lagrange completed a proof in 1775[14] based on his general theory of integral quadratic forms. The following presentation incorporates a slight simplification of his argument, due to Gauss, which appears in article 182 of the Disquisitiones Arithmeticae.

An (integral binary) quadratic form is an expression of the form ${\displaystyle ax^{2}+bxy+cy^{2}}$ with ${\displaystyle a,b,c}$ integers. A number ${\displaystyle n}$ is said to be represented by the form if there exist integers ${\displaystyle x,y}$ such that ${\displaystyle n=ax^{2}+bxy+cy^{2}}$. Fermat's theorem on sums of two squares is then equivalent to the statement that a prime ${\displaystyle p}$ is represented by the form ${\displaystyle x^{2}+y^{2}}$ (i.e., ${\displaystyle a=c=1}$, ${\displaystyle b=0}$) exactly when ${\displaystyle p}$ is congruent to ${\displaystyle 1}$ modulo ${\displaystyle 4}$.

The discriminant of the quadratic form is defined to be ${\displaystyle b^{2}-4ac}$. The discriminant of ${\displaystyle x^{2}+y^{2}}$ is then equal to ${\displaystyle -4}$.

Two forms ${\displaystyle ax^{2}+bxy+cy^{2}}$ and ${\displaystyle a'x'^{2}+b'x'y'+c'y'^{2}}$ are equivalent if and only if there exist substitutions with integer coefficients

${\displaystyle x=\alpha x'+\beta y'}$

${\displaystyle y=\gamma x'+\delta y'}$

with ${\displaystyle \alpha \delta -\beta \gamma =\pm 1}$ such that, when substituted into the first form, yield the second. Equivalent forms are readily seen to have the same discriminant, and hence also the same parity for the middle coefficient ${\displaystyle b}$, which coincides with the parity of the discriminant. Moreover, it is clear that equivalent forms will represent exactly the same integers, because these kind of substitutions can be reversed by substitutions of the same kind.

Lagrange proved that all positive definite forms of discriminant −4 are equivalent. Thus, to prove Fermat's theorem it is enough to find any positive definite form of discriminant −4 that represents ${\displaystyle p}$. For example, one can use a form

${\displaystyle px^{2}+2mxy+\left({\frac {m^{2}+1}{p}}\right)y^{2},}$

where the first coefficient a = ${\displaystyle p}$ was chosen so that the form represents ${\displaystyle p}$ by setting x = 1, and y = 0, the coefficient b = 2m is an arbitrary even number (as it must be, to get an even discriminant), and finally ${\displaystyle c={\frac {m^{2}+1}{p}}}$ is chosen so that the discriminant ${\displaystyle b^{2}-4ac=4m^{2}-4pc}$ is equal to −4, which guarantees that the form is indeed equivalent to ${\displaystyle x^{2}+y^{2}}$. Of course, the coefficient ${\displaystyle c={\frac {m^{2}+1}{p}}}$ must be an integer, so the problem is reduced to finding some integer m such that ${\displaystyle p}$ divides ${\displaystyle m^{2}+1}$: or in other words, a 'square root of -1 modulo ${\displaystyle p}$' .

We claim such a square root of ${\displaystyle -1}$ is given by ${\displaystyle K=\prod _{k=1}^{\frac {p-1}{2}}k}$. Firstly it follows from Euclid's Fundamental Theorem of Arithmetic that ${\displaystyle ab\equiv 0{\pmod {p}}\iff a\equiv 0{\pmod {p}}\ \ {\hbox{or}}\ \ b\equiv 0{\pmod {p}}}$. Consequently, ${\displaystyle a^{2}\equiv 1{\pmod {p}}\iff a\equiv \pm 1{\pmod {p}}}$: that is, ${\displaystyle \pm 1}$ are their own inverses modulo ${\displaystyle p}$ and this property is unique to them. It then follows from the validity of Euclidean division in the integers, and the fact that ${\displaystyle p}$ is prime, that for every ${\displaystyle 2\leq a\leq p-2}$ the gcd of ${\displaystyle a}$ and ${\displaystyle p}$ may be expressed via the Euclidean algorithm yielding a unique and distinct inverse ${\displaystyle a^{-1}\neq a}$ of ${\displaystyle a}$ modulo ${\displaystyle p}$. In particular therefore the product of all non-zero residues modulo ${\displaystyle p}$ is ${\displaystyle -1}$. Let ${\displaystyle L=\prod _{l={\frac {p+1}{2}}}^{p-1}l}$: from what has just been observed, ${\displaystyle KL\equiv -1{\pmod {p}}}$. But by definition, since each term in ${\displaystyle K}$ may be paired with its negative in ${\displaystyle L}$, ${\displaystyle L=(-1)^{\frac {p-1}{2}}K}$, which since ${\displaystyle p}$ is odd shows that ${\displaystyle K^{2}\equiv -1{\pmod {p}}\iff p\equiv 1{\pmod {4}}}$, as required.

Richard Dedekind gave at least two proofs of Fermat's theorem on sums of two squares, both using the arithmetical properties of the Gaussian integers, which are numbers of the form a + bi, where a and b are integers, and i is the square root of −1. One appears in section 27 of his exposition of ideals published in 1877; the second appeared in Supplement XI to Peter Gustav Lejeune Dirichlet's Vorlesungen über Zahlentheorie, and was published in 1894.

1. First proof. If ${\displaystyle p}$ is an odd prime number, then we have ${\displaystyle i^{p-1}=(-1)^{\frac {p-1}{2}}}$ in the Gaussian integers. Consequently, writing a Gaussian integer ω = x + iy with x,y ∈ Z and applying the Frobenius automorphism in Z[i]/(p), one finds

${\displaystyle \omega ^{p}=(x+yi)^{p}\equiv x^{p}+y^{p}i^{p}\equiv x+(-1)^{\frac {p-1}{2}}yi{\pmod {p}},}$

since the automorphism fixes the elements of Z/(p). In the current case, ${\displaystyle p=4n+1}$ for some integer n, and so in the above expression for ω^p, the exponent (p-1)/2 of -1 is even. Hence the right hand side equals ω, so in this case the Frobenius endomorphism of Z[i]/(p) is the identity.

Kummer had already established that if f ∈ {1,2} is the order of the Frobenius automorphism of Z[i]/(p), then the ideal ${\displaystyle (p)}$ in Z[i] would be a product of 2/f distinct prime ideals. (In fact, Kummer had established a much more general result for any extension of Z obtained by adjoining a primitive m-th root of unity, where m was any positive integer; this is the case m = 4 of that result.) Therefore, the ideal (p) is the product of two different prime ideals in Z[i]. Since the Gaussian integers are a Euclidean domain for the norm function ${\displaystyle N(x+iy)=x^{2}+y^{2}}$, every ideal is principal and generated by a nonzero element of the ideal of minimal norm. Since the norm is multiplicative, the norm of a generator ${\displaystyle \alpha }$ of one of the ideal factors of (p) must be a strict divisor of ${\displaystyle N(p)=p^{2}}$, so that we must have ${\displaystyle p=N(\alpha )=N(a+bi)=a^{2}+b^{2}}$, which gives Fermat's theorem.

2. Second proof. This proof builds on Lagrange's result that if ${\displaystyle p=4n+1}$ is a prime number, then there must be an integer m such that ${\displaystyle m^{2}+1}$ is divisible by p (we can also see this by Euler's criterion); it also uses the fact that the Gaussian integers are a unique factorization domain (because they are a Euclidean domain). Since p ∈ Z does not divide either of the Gaussian integers ${\displaystyle m+i}$ and ${\displaystyle m-i}$ (as it does not divide their imaginary parts), but it does divide their product ${\displaystyle m^{2}+1}$, it follows that ${\displaystyle p}$ cannot be a prime element in the Gaussian integers. We must therefore have a nontrivial factorization of p in the Gaussian integers, which in view of the norm can have only two factors (since the norm is multiplicative, and ${\displaystyle p^{2}=N(p)}$, there can only be up to two factors of p), so it must be of the form ${\displaystyle p=(x+yi)(x-yi)}$ for some integers ${\displaystyle x}$ and ${\displaystyle y}$. This immediately yields that ${\displaystyle p=x^{2}+y^{2}}$.

For ${\displaystyle p}$ congruent to ${\displaystyle 1}$ mod ${\displaystyle 4}$ a prime, ${\displaystyle -1}$ is a quadratic residue mod ${\displaystyle p}$ by Euler's criterion. Therefore, there exists an integer ${\displaystyle m}$ such that ${\displaystyle p}$ divides ${\displaystyle m^{2}+1}$. Let ${\displaystyle {\hat {i}},{\hat {j}}}$ be the standard basis elements for the vector space ${\displaystyle \mathbb {R} ^{2}}$ and set ${\displaystyle {\vec {u}}={\hat {i}}+m{\hat {j}}}$ and ${\displaystyle {\vec {v}}=0{\hat {i}}+p{\hat {j}}}$. Consider the lattice ${\displaystyle S=\{a{\vec {u}}+b{\vec {v}}\mid a,b\in \mathbb {Z} \}}$. If ${\displaystyle {\vec {w}}=a{\vec {u}}+b{\vec {v}}=a{\hat {i}}+(am+bp){\hat {j}}\in S}$ then ${\displaystyle \|{\vec {w}}\|^{2}\equiv a^{2}+(am+bp)^{2}\equiv a^{2}(1+m^{2})\equiv 0{\pmod {p}}}$. Thus ${\displaystyle p}$ divides ${\displaystyle \|{\vec {w}}\|^{2}}$ for any ${\displaystyle {\vec {w}}\in S}$.

The area of the fundamental parallelogram of the lattice is ${\displaystyle p}$. The area of the open disk, ${\displaystyle D}$, of radius ${\displaystyle {\sqrt {2p}}}$ centered around the origin is ${\displaystyle 2\pi p>4p}$. Furthermore, ${\displaystyle D}$ is convex and symmetrical about the origin. Therefore, by Minkowski's theorem there exists a nonzero vector ${\displaystyle {\vec {w}}\in S}$ such that ${\displaystyle {\vec {w}}\in D}$. Both ${\displaystyle \|{\vec {w}}\|^{2}<2p}$ and ${\displaystyle p\mid \|{\vec {w}}\|^{2}}$ so ${\displaystyle p=\|{\vec {w}}\|^{2}}$. Hence ${\displaystyle p}$ is the sum of the squares of the components of ${\displaystyle {\vec {w}}}$.

Let ${\displaystyle p=4k+1}$ be prime, let ${\displaystyle \mathbb {N} }$ denote the natural numbers (with or without zero), and consider the finite set ${\displaystyle S=\{(x,y,z)\in \mathbb {N} ^{3}:x^{2}+4yz=p\}}$ of triples of numbers. Then ${\displaystyle S}$ has two involutions: an obvious one ${\displaystyle (x,y,z)\mapsto (x,z,y)}$ whose fixed points ${\displaystyle (x,y,y)}$ correspond to representations of ${\displaystyle p}$ as a sum of two squares, and a more complicated one,

${\displaystyle (x,y,z)\mapsto {\begin{cases}(x+2z,~z,~y-x-z),\quad {\textrm {if}}\,\,\,x<y-z\\(2y-x,~y,~x-y+z),\quad {\textrm {if}}\,\,\,y-z<x<2y\\(x-2y,~x-y+z,~y),\quad {\textrm {if}}\,\,\,x>2y\end{cases}}}$

which has exactly one fixed point ${\displaystyle (1,1,k)}$. This proves that the cardinality of ${\displaystyle S}$ is odd. Hence, ${\displaystyle S}$ has also a fixed point with respect to the obvious involution.

This proof, due to Zagier, is a simplification of an earlier proof by Heath-Brown, which in turn was inspired by a proof of Liouville. The technique of the proof is a combinatorial analogue of the topological principle that the Euler characteristics of a topological space with an involution and of its fixed-point set have the same parity and is reminiscent of the use of sign-reversing involutions in the proofs of combinatorial bijections.

This proof is equivalent to a geometric or "visual" proof using "windmill" figures, given by Alexander Spivak in 2006 and described in this MathOverflow post and this Mathologer YouTube video Why was this visual proof missed for 400 years? (Fermat's two square theorem) on YouTube.

In 2016, A. David Christopher gave a partition-theoretic proof by considering partitions of the odd prime ${\displaystyle n}$ having exactly two sizes ${\displaystyle a_{i}(i=1,2)}$, each occurring exactly ${\displaystyle a_{i}}$ times, and by showing that at least one such partition exists if ${\displaystyle n}$ is congruent to 1 modulo 4.[15]