Taken from problem 46 on Project Euler:
It was proposed by Christian Goldbach that every odd composite number
can be written as the sum of a prime and twice a square.$9 = 7 + 2 \times 1^2$
$15 = 7 + 2 \times 2^2$
$21 = 3 + 2 \times 3^2$
…
It turns out that the conjecture was false.
Now, suppose the phrase "twice a square" was replaced with "two squares." In other words, this revision of the conjecture would state that all odd composites $C$ can be written in the following form:
$C=p+a^2+b^2$
Where $p$ is a prime and $a$ and $b$ can take on any non-negative integer values. Making the substitution $a=b$ reduces this problem to the classical conjecture.
Now, is this weaker conjecture true?
My Attempt at the Problem:
Let $d=a^2+b^2$. A well-known theorem states that the sum of two squares can be rewritten as $2^{e_0} \left( {p_1}^{e_1}…{p_n}^{e_n} \right) \left( {q_1}^{2 o_1}…{q_n}^{2 o_n} \right)$, where $p_i$ is a prime of the form $4k+1$ and $q_i$ is a prime of the form $4k+3$. We can therefore write the conjecture as follows:
$C=p+d$
$C=p+ 2^{e_0} \left( {p_1}^{e_1}…{p_n}^{e_n} \right) \left( {q_1}^{2 o_1}…{q_n}^{2 o_n} \right)$
$C=p+ 2^{e_0}g$
Where $g$ is any integer such that $g=1 \left( \text{mod } 4 \right)$. Letting $p=2$ and $e_0=0$ shows us that the conjecture is true when $C$ is of the form $4k+3$.
From here I am not sure where to go. Any clues?
Best Answer
Yes, this conjecture is true. Moreover, Hua in $1938$ proved that if $n$ is odd and $n\ne 2 (\mod 3),$ then $n=p_1^2+p_2^2+p$ where $p_1$ and $p_2$ are primes. Unfortunately, all methods to prove such results are far from being elementary.