# Primes of the form : $x^2+4\cdot y^2$

How to prove following statement :

Conjecture:

Odd prime $p$ is expressible as : $p=x^2+4\cdot y^2$ , $x,y > 0$

if and only if : $p\equiv 1 \pmod {12}$ or $p\equiv 5 \pmod {12}$ .

Similar statements (without a proof) related to the Fermat's theorem on sums of two squares can be found here .

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## 4 Answers

This is an immediate consequence of Fermat's theorem that an odd prime $p$ is a sum of two squares if and only if $p \equiv 1$ (mod 4). Such a prime $p$ is congruent to $1$ (mod 3) or $2$ (mod 3). In the former case, $p \equiv 1$ (mod 12), and in the latter case, $p \equiv 5$ (mod 12). Given a prime $p \equiv 1$ (mod 4), express it in the form $a^2 + b^2,$ for positive integers $a$ and $b.$ Now $a$ and $b$ can't both be of the same parity, so assume that $a$ is even and $b$ is odd. If $ab$ is not divisible by $3,$ we have $p \equiv 2$ (mod 3), so $p \equiv 5$ (mod 12). If $ab$ is divisible by $3,$ we have $p \equiv 1$ (mod 3) and so $p \equiv 1$ (mod 12).

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First to prove for the "only if" clause in a similar method to one found here:

For squares modulo 12 we have {0, 1, 4, 9} as the only possibilities for any $x^2$ (Quadratic residues), while only {0, 4} is valid for any $4\cdot y^2 = (2y)^2$. For primes $p > 12$ modulo 12 we have {1, 5, 7, 11}, the integers coprime to 12, as the only possibilities. Trying all possible options will show that 7 and 11 (mod 12) cannot be formed no matter how you pick the squares, while 1 and 5 are possible due to 1 + 0 (mod 12) and 1 + 4 (mod 12) respectively.

Checking for the remaining primes $p <= 12$ can then be done by hand.

Now noting that both 1 and 5 modulo 12 are equivalent to 1 modulo 4 allows us to apply the very theorem you linked in your question for the "is expressible if" clause, the proof of which is above on Wikipedia.

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Modulo $12$, the quadratic residues are:

$[1,4,9,4,1,0,1,4,9,4,1]$. Of these, odd residues are $1,9$.

And the residues of the form $4\times i^2 \pmod{12}$ are of the form

$[4,4,0,4,4,0,4,4,0,4,4]$

Hence, any $\textbf{odd}$ number of the form $(j^2+4\times i^2)$ has to be either $1$ or $5\pmod{12}$, regardless of whether or not it is a prime.

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HINT $\rm\ (\Rightarrow)\ \ odd\ p\ =\ x^2+4\ y^2\ \Rightarrow\ x\ odd,\ so\ \ mod\ 4\!:\ x\ \equiv\: \pm1\ \Rightarrow\ p\ \equiv\ x^2\:\equiv\ 1$

Thus $\rm\ p\equiv 1\pmod 4\ \Rightarrow\ p\equiv 1,5,9\pmod{12}\:,\:$ but not $\:9\:$ since $\rm\:3\:|\:p\ \Rightarrow\ p = 3\ne x^2 + 4\ y^2$

$\rm(\Leftarrow)\$ By Fermat $\rm\: p \equiv 1\pmod 4\: \Rightarrow\ p\: =\: m^2 + n^2\: \Rightarrow\ 2\ |\ m\ \ or\ \:n\ ($else $\rm\: 2\ |\ p = odd^2+odd^2)$

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