Find if there exists an element which is sum of squares

Question

Consider the sequence : $\langle 3,7,11,15,\cdots\rangle$. Is it possible that an element of this sequence is sum of two squares?

My approach:

General term of the sequence if $4n-1$. Consider the canonical homomorphism $\phi \colon \mathbb{Z} \to \mathbb{Z}_4$. If $a^2+b^2=4n-1$ has a solution in $\mathbb{Z}$, then it has a solution in $\mathbb{Z}_4$. That is, $$a^2+b^2=(4n-1) \ \text{mod}\ 4=3 $$

Clearly, this has no solution in $\mathbb{Z}_4$ and hence neither does the original.

What I would like to know is if there are any other methods to solve this question using maybe some other areas of mathematics?

Answer 1

Are you fine with algebraic number theory? The ring of gaussian integers $\mathbb{Z}[i]$ is a euclidean domain, hence a unique factorization domain. $n=a^2+b^2$ is equivalent to $n= z\cdot\overline{z}$ with $z=(a+ib)\in\mathbb{Z}[i]$, hence the problem of understanding which positive integers can be represented as a sum of two squares boils down to understanding which integer primes can be represented as a sum of two squares. Lagrange's identity $$(a^2+b^2)(c^2+d^2) = (ad-bc)^2 + (ac+bd)^2 $$ is a consequence of $(a-ib)(c+id) = (ac+bd)+i(ad-bc)$ and grants that the set of integers that can be represented as a sum of two squares is a semigroup.
If $p\in\mathbb{Z}^+$ is an odd prime and $a^2+b^2=p$, we have that both $a$ and $b$ are invertible elements in $\mathbb{Z}/(p\mathbb{Z})$ and the square of $a^{-1} b$ equals $-1$. In particular, if $p$ is an odd prime that can be represented as a sum of two squares, $-1$ is a quadratic residue in $\mathbb{Z}/(p\mathbb{Z})^*$. By Legendre's symbol $$ 1 = \left(\frac{-1}{p}\right) \equiv (-1)^{\frac{p-1}{2}} \pmod{p}$$ we get that $-1$ is a quadratic residue in $\mathbb{Z}/(p\mathbb{Z})^*$ iff $\frac{p-1}{2}$ is even, i.e. iff $p=4k+1$.

Corollary1. Every prime of the form $4k+1$ can be represented as the sum of two squares and no prime of the form $4k+3$ can be represented as the sum of two squares.

Corollary2. Let $n\in\mathbb{Z}^+$ and let $$ n = 2^{k} p_1^{\alpha_1}\cdots p_{l}^{\alpha_l}\cdot q_1^{\beta_1}\cdots q_m^{\beta_m} $$ be the factorization of $n$, with $p_i\equiv 1\pmod{4}$ and $q_j\equiv 3\pmod{4}$. $n$ can be represented as a sum of two squares iff $\beta_1\equiv \ldots\equiv \beta_m\equiv 0\pmod{2}$.

Corollary3. Every positive integer of the form $4k+3$ has a prime divisor $p$ of the same form, such that $\nu_p(n)$ is odd. In particular, no positive integer of the form $4k+3$ can be represented as the sum of two squares.

Isn't this the most epic twist on a trivial argument you have ever seen? :D