Primes with Special Squares

We want to find integers $a,b,p$ such that $a^3 \; = \; p^2 + b^2 \; = \; (p + ib)(p - ib)$ where $p$ is prime and $a,b$ are coprime. Working the in Euclidean domain $\mathbb{Z}[i]$ of Gaussian integers, consider the irreducible factors of $a$ .

• No integer $u \in \mathbb{Z}$ , with $u > 1$ , can be a factor of $p+ib$ . If such a $u$ existed, then it would have to divide $p$ in $\mathbb{Z}$ , and hence would be equal to $p$ . But this would imply that $p$ divided $b$ , and hence that $p$ divided $a$ , which would mean that $a$ and $b$ were not coprime.

• No irreducible element $\zeta$ of $\mathbb{Z}[i]$ exists such that both $\zeta$ and its complex conjugate $\zeta^\star$ divide $p + ib$ . If such a $\zeta$ existed, then $|\zeta|^2$ would be a nontrivial integer factor of $p + ib$ .

If $\zeta$ is an irreducible factor of $a$ , then it divides exactly one of $p+ib$ and $p-ib$ (it cannot divide both, for that would make $\zeta$ and $\zeta^\star$ both divide $p+ib$ ). Since $a^3 = (p+ib)(p-ib)$ , if the index of $\zeta$ in $a$ is $\alpha$ then the index of $\zeta$ in either $p+ib$ or $p-ib$ must be $3\alpha$ , a multiple of $3$ . Thus we deduce that $p+ib$ must be a product of cubes of irreducible factors of $\mathbb{Z}[i]$ , none of which are integers, and hence we deduce that $p + ib \; = \; (x + iy)^3$ for some integers $x,y$ . This tells us that $\begin{array}{rcl} a & = & x^2 + y^2 \\ b & = & 3x^2y - y^3 \\ p & = & x^3 - 3xy^2 \; = \; x(x^2 - 3y^2) \end{array}$ There are now four possibilities for $x$ :

1. If $x=p$ , then $1 = p^2 - 3y^2$ , and hence $y^2 \,=\, \tfrac13(p^2-1)$ . There are only three primes less than $300$ which have this property, and we have the following table of corresponding values: $\begin{array}{|c|c|c|c|c|} \hline p & x & y & a & b \\ \hline 2 & 2 & 1 & 5 & 11 \\ 7 & 7 & 4 & 65 & 524 \\ 97 & 97 & 56 & 12545 & 1405096 \\ \hline \end{array}$ and $a$ and $b$ are coprime in all three cases.

2. If $x=1$ , then $p = 1 - 3y^2$ , and hence $y^2 \,=\, \tfrac13(1-p) \,<\, 0$ , which is not possible.

3. If $x=-1$ , then $-p =1 - 3y^2$ , and hence $y^2 \,=\, \tfrac13(1+p)$ . There are five primes less than $300$ which have this property, and we have the following table of corresponding values: $\begin{array}{|c|c|c|c|c|} \hline p & x & y & a & b \\ \hline 2 & -1 & 1 & 2 & 2 \\ 11 & -1 & 2 & 5 & -2 \\ 47 & -1 & 4 & 17 & -52 \\ 107 & -1 & 6 & 37 & -198 \\ 191 & -1 & 8 & 65 & -488 \\ \hline \end{array}$ and $a$ and $b$ are coprime in all but the first case. Happily, the prime $2$ has already qualified.

4. If $x=-p$ , then $-1 = p^2 - 3y^2$ , and hence $y^2 \,=\, \tfrac13(p^2+1)$ . But $n^2+1$ is never a multiple of $3$ .

Thus there are $7$ primes less than $300$ for which this decomposition can be made: $2,7,97,11,47,107,191$ . Their sum is $462$ .

The main trick is to rewrite the identity $p^2 = a^3 - b^2$ in the form $a^3 = p^2 + b^2$ . The sum of squares then should remind you of the Gaussian integers!

First, $2^2 = 5^3 - 11^2.$ Henceforth, we can assume $p$ is odd. By looking mod $4$ , we can see that in this case, $b$ is even and $a$ is odd.

Write $(p+bi)(p-bi) = a^3$ . Note that $p+bi$ and $p - bi$ are relatively prime in the Gaussian integers (this makes sense since the Gaussian integers are a Euclidean domain): any common divisor divides their sum and difference, hence $2p$ and $2b$ , but $p|b$ is impossible because then $p|a$ , contradicting coprimality; so any common divisor divides $2$ , but $p+bi$ has odd norm and norms are multiplicative, so nothing with even norm can divide it.

When the product of two relatively prime elements is a cube, they must both be unit multiples of cubes; but all the units are cubes themselves, so $p+bi$ and $p-bi$ are cubes. Write $p+bi = (x+yi)^3.$ (Note that this will imply $p-bi = (x-yi)^3$ and hence $a = (x^2+y^2)$ . Equating real and imaginary parts gives $p = x(x^2-3y^2).$ There are two cases: $x = \pm 1$ , $x = \pm p$ .

If $x = \pm 1$ , we get $1-3y^2 = \pm p.$ Clearly this must be $-p$ , hence $p = 3y^2 -1$ . Note that $y$ must be even to give odd $p$ . Looking at even values of $y \le 10$ gives $p = 11, 47, 107, 191, 299$ ; but $299$ is not prime, so we're left with $11, 47, 107, 191.$

If $x = \pm p$ , we get $p^2 - 3y^2 = \pm 1.$ Looking mod $3$ we must have $p^2 - 3y^2 = 1$ . The positive solutions to this Pell equation can be enumerated in any number of ways, for instance by starting with $a_1 = 2, b_1 = 1$ and then using the recursive formulas $a_{n+1} = 2a_n + 3b_n, b_{n+1} = a_n + 2b_n.$ This gives $(2,1), (7,4), (26,15), (97,56), (362,209), \ldots$ , from which we take the viable solutions $7, 97$ .

It is not hard to go through each of the equations once $x$ and $y$ are determined, to solve for $a$ and $b$ , to check that each of these yield solutions.

Summing up, we get $2 + 11 + 47 + 107 + 191 + 7 + 97 = \fbox{462}$ .

[Note: The case $p=2$ is dealt with seperately. Do you see where in this prove we need $p\neq 2$ ? - Calvin] (Some high-power tools are involved) We have $a^3 = p^2 + b^2$ . It is easy to note that p,a,b must be pairwise coprime (otherwise, if p divides a or b, then it must divide the other). Factoring in $\mathbb{Z}[i]$ , we have that $a^3 = (p+bi)(p-bi)$ . However, $\gcd(p+bi,p-bi) = \gcd(p+bi, 2bi) = \gcd(p+bi, bi)$ , since 2 doesn't divide $p+bi$ as $p,b$ are coprime). This is thus equal to $\gcd(p, bi) = 1.$ Thus we have that for some $k=x+iy$ , $k^3 = p+bi$ , $\overline{k} ^3 = p-bi$ .

Now we expand the first equation: $p+bi = (x+iy)^3 = x(x^2-3y^2) + iy(3x^2 - y^2)$ Thus $p = x(x^2-3y^2)$ . This implies $x=1$ or -1, or that $x^2 - 3y^2 = 1$ or -1. We have the following cases

If $x = 1$ , then $p = 1-3y^2 < 2$ , contradiction

If $x = -1$ , then $p = 3y^2 - 1$ . We may now list a few such values to obtain $(y,p) = (1,2) (2,11) (4,47) (6,107) (8,191)$ However, taking $(y,p) = 1,2$ will not give coprime values of $p$ and $b$ since $b = 2$ . It is simple to verify that the other solutions are valid.

If $x^2-3y^2 = -1$ , we note that there are no integer solutions by taking mod 4, which is equivalent to $x^2 + y^2 \equiv -1 \pmod{4}$ . Of course, we know that $x^2 +y^2$ can only be congruent to 0,1,2 mod 4.

If $x^2 - 3y^2 = 1$ , then we have a Pell's equation (,as well as $p = x$ ). The fundamental solution is $x=2, y=1$ . Then, we expand $(2+\sqrt{3})^n$ to get solutions: $(x,y) = (2,1) (7,4), (97, 56)$

This gives us the last three solutions, and we may easily check that these solutions are also valid. Thus $2+7+11+47+97+107+191 = 462.$