Prime Diophantine!

Consider the the equation modulo $3$ . Since perfect squares are $\equiv 0, 1 \pmod 3$ , it follows that $3 \mid m$ and $3 \mid p^2 n^2$ .

Now we divide the whole thing into two cases.

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Case 1 (when $p \neq 3$ )

This immediately forces $3\mid n$ .

Now note that left hand side is divisible by $9$ while the right hand side is not.

This means there is no solution when $p\neq 3$ .

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Case 2 (when $p =3$ )

In this case, our equation transforms into

$m^2-3mn+9n^2=36$

This equation is much easier to work with since now we have two unknowns instead of three.

Remember that $3\mid m$ . So we can divide both sides by $9$ to get an equation that looks like this.

${m_1}^2-m_1 n +n^2=4$

where $m_1$ is an integer such that $3m_1=m$ .

Take a look at the equation again.

It is apparent that both $m_1$ and $n$ have to be even. Even if one of them is odd, this equation can not hold.

Divide both sides of the equation by $4$ to obtain:

${m_2}^2-m_2n_2+{n_2}^2=1$

where $m_2$ and $n_2$ are integers satisfying $m_1=2m_2$ and $n=2n_2$ .

Now this has nothing to do with this problem, but note that this can be written as $\dfrac{{m_2}^2+{n_2}^2}{m_2n_2+1}=1$ .

Doesn't this look familiar?

Well, it is the infamous IMO 1988/6 with the right hand side equal to $1$ .

Anyway, we're almost done. Just use the AM-GM inequality to see that ${m_2}^2+{n_2}^2 \geq |2m_2n_2| \geq |m_2n_2+1|$ when $m_2$ and $n_2$ are integers.

In other words, the numerator grows much faster than the denominator.

The only solutions are $(1,1), (-1, -1), (0, 1), (1, 0), (-1, 0)$ and $(0 -1)$ .

So the only solutions to the original equation are $(6, 2, 3), (-6, -2, 3), (0, 2, 3), (6, 0, 3), (-6, 0, 3)$ and $0, -2, 3)$ .

We have a total of $\boxed{6}$ solutions.

When $p\neq 3$ we have $p^2\equiv 1\pmod 3$ so taking mod $3$ reveals

$m^2+n^2\equiv 0\pmod 3\implies m\equiv n\equiv 0\pmod 3$

since quadratic residues mod $3$ are only $0$ and $1$ . Setting $m=3m_0$ and $n=3n_0$ we have

$9m_0^2-27m_0n_0+9p^2n_0^2=12p\implies 9\mid 3p\implies p=3$

a contradiction, so $p=3$ . The equation is a quadratic in $m$ with discriminant $48p+9n^2-4n^2p^2=144-27n^2\ge 0\implies |n|\le \dfrac{4}{\sqrt 3}$ and since $n$ is an integer, $|n|\le 2$ . It's easy to see that both of $m$ and $n$ are even. So checking all $n\in\{-2,0,2\}$ we find $\fbox{6}$ sets of solution:

$(m,n,p)=(0,-2,3),(-6,-2,3),(6,0,3),(-6,0,3),(0,2,3),(6,2,3).$

My solution is quite long, but here it goes.

$m^{ 2 }-3mn+p^{ 2 }n^{ 2 }=12p$ $\Leftrightarrow 4m^{ 2 }-12mn+4p^{ 2 }n^{ 2 }=48p$ $\Leftrightarrow 4m^{ 2 }-12mn+9n^{ 2 }+n^{ 2 }(4p^{ 2 }-9)=48p$ $\Leftrightarrow (2m-3n)^{ 2 }+n^{ 2 }(4p^{ 2 }-9)=48p$ Since $(2m-3n)^{ 2 } \ge 0$ , we need $n^{ 2 }(4p^{ 2 }-9)\le 48p$ , or $4p^{ 2 }-9\le 48p$ since $n^2\ge 0$ . Solve the inequality gives $\dfrac { -\sqrt { 153 } +12 }{ 2 } \le p\le \dfrac { \sqrt { 153 } +12 }{ 2 }$ .

Since $p$ is a prime, $p$ can only be $2$ , $3$ , $5$ , $7$ or $11$ . Replacing each value of $p$ into the first Diophantine equation gives us a total of $6$ triplets of $(m,n,p)$ , which are $(6,0,3),(-6,0,3),(6,2,3),(0,2,3),(0,-2,3),(-6,-2,3)$ .

WolframAlpha {m^2 - 3mn + (p^2 n^2) -12p =0, p>0}

m, n, p = (+/- 6, 0, 3) or (0, +/-2, 3) or (-6, -2, 3) or (6, 2, 3)

6 solutions