# Diophantine equation: $n^p+3^p=k^2$

Find all solutions to the Diophantine equation $n^p+3^p=k^2$, where $p\in \mathbb{P}$ and $n,k$ positive integers.

I have tried everything, from mods to bounding to LTE; nothing seems to work on this. I did find one solution: $(n,p,k)=(4,2,5)$, which was motivated by noticing the resemblance to Pythagorean triples.

I should note that I don't know any advanced number theory (I'm in high school), so I apologize if there is a very simple approach I'm not seeing.

• Technically, not called a Diophantine equation, because of the variable in the exponent. Sometimes called an exponential Diophantine equation... – Thomas Andrews Jul 30 '14 at 15:45
• Where did you encounter this problem? – abiessu Jul 30 '14 at 15:47
• Is there any evidence that the poser knows the solution? – Thomas Andrews Jul 30 '14 at 15:54
• Looking at this in $\mathbb{Z} / p\mathbb{Z}$, you find that necessarily $k^2 = n+3$. Finding the square root in $\mathbb{Z} / p\mathbb{Z}$ is to my knowledge very difficult. – Matt B. Jul 30 '14 at 16:25
• Where in AoPS is this post? – Mayank Pandey Sep 29 '14 at 1:43

This is not a real answer but a summary of what I know for $p > 3$.

In addition to being called exponential Diophantine equation, the equation $$n^p + 3^p = k^2\tag{*1}$$ is a special case of something called Super Fermat equation:

$$x^p + y^q = z^r\quad\text{ where }\quad x, y, z \in \mathbb{Z}\;\;\text{ and }\;\; p, q, r \ge 2,\; \max(p,q,r) > 2$$

A solution of it is called trivial if $xyz = 0$ or $\pm 1$ and called primitive if $\gcd(x,y,z) = 1$.

In 2001, in a paper Winding quotients and some variants of Fermat’s Last Theorem,
Darmon and Merel has shown

The equation $x^s + y^s = z^2$ has no non-trivial primitive solution when $s > 3$

If we apply this to our equation $n^p + 3^p = k^2$. We find when $p > 3$, it cannot have any solution unless $\gcd(n, 3, k ) = 3$. Writing $n$ as $3x$, we have

$$3^{p} (x^p + 1) = k^2 \quad\implies\quad 3^{\lceil p/2 \rceil} | k$$

Writing $k$ as $3^{\lceil p/2\rceil}z$ and notice by assumption, $p$ is an odd prime, $(*1)$ reduces to

$$x^p + 1 = 3z^2\tag{*2}$$

I have no idea whether this equation has a solution at all.

However, in another paper One the equations $z^m = F(x,y)$ and $Ax^p + By^q = Cz^r$ , Darmon and Graville has shown:

Suppose $1/p + 1/q + 1/r < 1$ and $A, B, C \in \mathbb{Z}$ with $ABC \ne 0$. Then the number of solutions to the equation $$A x^p + By^q = C z^r,\quad\gcd(x,y,z) = 1\tag{*3}$$ is finite.

In $(*3)$, if we

• pull the values of $x$ and $p$ from $(*2)$.
• set $A, B, C, y, r$ to $1, 1, 3, 1, 2$ respectively,
• pick a $q$ so large such that $1/p + 1/q + 1/r < 1$,

we will find $\gcd(x,y,z) = \gcd(x,1,z) = 1$ and hence we can use above theorem to conclude:

The equation $n^p + 3^p = k^2$ has at most finitely many solutions for $p > 3$.
If such a solution exists, it must have the form $n = 3x$ and $k = 3^{\lceil p/2 \rceil} y$.

• Just a note : Intuitively $(*3)$ is almost surely true. From a heuristic point of view, there are $O(n^{1/p})$ $p$-powers, $O(n^{1/q})$ $q$-powers and $O(n^{1/r})$ $r$-powers upto some integer $n$ so any additive relation is expected to happen $O(n^{1/p} \cdot n^{1/q} \cdot n^{1/r})$ times. If $1/p + 1/q + 1/r < 1$ then one should expect that the number of solutions as $n \to \infty$ is finite. Good answer, by the way. +1. – Balarka Sen Jul 30 '14 at 19:59

There are no more solutions with $p = 2,$ apart form $n = 4, k= 5.$ To see this note that the general positive integer solution to $x^{2} +y^{2} = z^{2}$ with $x$ and $y$ having no common factor greater than $1$ is parametrized in the form $x = a^{2}-b^{2}, y = 2ab$ and $z = a^{2}+b^{2},$ where $a>b$ are positive integers with no common factor greater than $1$ and $a$ and $b$ are not both odd. The only way to get $x = 3$ with such $a,b$ is with $a-b = 1, a+b = 3,$ so $a =2,b=1,$ yielding $y =4, z = 5.$

I do not know if there any solutions with odd primes $p,$ but I know what form they would have to take if there were any: there are two general forms with $n$ and $3$ having no common factor greater than $1$: if $p$ divides $n+3,$ there must be some integers $r,s$ and $t$ such that $n + 3 = p^{2r-1}s^{2}$ and $pt ^{2} = \sum_{j=0}^{p-1} n^{p-1-j} (-3)^{j}.$ But if $p$ does not divide $n+3,$ there must be integers $s$ and $t$ such that $n + 3 = s^{2}$ and $t ^{2} = \sum_{j=0}^{p-1} n^{p-1-j} (-3)^{j}.$

I won't give all details, but the reasoning is the same as leads to the two cases in Fermat's Last Theorem $p |xyz$ and $p \not | xyz.$ However, in this case, I am not sure that knowing the form of the solution helps. Perhaps there is some easy argument in this case, but I don't see it at the moment.