Exponential growth

This doesn't require any fancy theorems.

Write $p^x = (y+1)(y^2-y+1)$ , and note that the gcd of $y+1$ and $y^2-y+1$ is $1$ or $3$ , because $y^2-y+1 = (y+1)(y-2)+3$ . Both factors are powers of $p$ , so their gcd is the minimum of them. This restricts the possibilities for them quite dramatically.

If the gcd is $1$ then one of the factors has to be $1$ . This is only possible if $y = 1$ , which leads to $(2,1,1)$ .

If the gcd is $3$ then $p = 3$ and one of the factors is $3$ . The only possibility is $y = 2$ (whence both factors are $3$ ) which leads to $(3,2,2)$ .

That's it.

This equation can be rewritten as $p^{x} = y^{3} + 1 = (y + 1)(y^{2} - y + 1),$ (A).

Thus by the Fundamental Theorem of Algebra we must have that

$y + 1 = p^{m}$ and $y^{2} - y + 1 = p^{n}$ for some non-negative integers $m,n$ with $x = m + n.$

Now looking again at equation (A), we see that since $p,x,y \gt 1$ we can have at most one of $m,n$ be zero. So we must look at the following three cases:

• (i) $m = 0$ : then $p^{n} = p^{0} - 3(p^{0} - 1) = 1,$ which can be discarded since $p \gt 1;$

• (ii) $n = 0$ : then $p^{0} = (p^{m})^{2} - 3(p^{m} - 1) \Longrightarrow (p^{m})^{2} - 3p^{m} + 2 = 0$

$\Longrightarrow (p^{m} - 2)(p^{m} - 1) = 0 \Longrightarrow p^{m} = 2$ or $p^{m} = 1.$

Given these options, the only valid solution is $p = 2, m = 1.$ This gives us $x = m + n = 1, y = 1$ for the solution triple $(x,y,p) = (1,1,2).$

• (iii) $m \ne 0, n \ne 0$ : In this case $p$ divides both $y + 1$ and $y^{2} - y + 1.$ Thus $p$ divides $(y^{2} - y + 1) - (y + 1) = y(y + 2).$ But since $p | (y + 1)$ we know that $p$ does not divide $y.$ Thus $p | (y - 2),$ which in turn implies that $p$ divides $(y + 1) - (y - 2) = 3,$ and the only prime that divides $3$ is itself.

Now $y = 1$ gives us $3^{x} - 1 = 1,$ which has no integer solution. With $y = 2$ we have $3^{x} - 8 = 1,$ giving us the solution triple $(x,y,p) = (2,2,3).$

Next, suppose $y \ge 3.$ Now we'll make use of Catalan's Conjecture , (which was proven in 2002), which states that the only pair of "non-trivial" solution of the equation $a^{x} - y^{b} = 1$ is $3^{2} - 2^{3} = 1.$ By "non-trivial" it is meant that the powers are $\ge 2,$ and since $x = m + n \ge 2$ the Conjecture applies in this case.

Thus having covered all the cases, we find that the only solution triples are $(1,1,2)$ and $(2,2,3),$ so the desired sum is $(1 + 1 + 2) + (2 + 2 + 3) = \boxed{11}.$

Clearly $y\ge 1$ and $p^x=y^3+1$ . Zsigmondy tells us this lemma:

$a^n+b^n$ has at least two prime divisors when $a>b>0$ , $n\ge 2$ , $(a,b,n)\neq (2,1,3)$ .

$(a,b,n)=(2,1,3)$ here gives $\boxed{(x,y,p)=(2,2,3)}$ . Otherwise, since $3\ge 2$ , $y>1>0$ will give contradiction by above lemma. So $y=1$ , which gives $\boxed{(x,y,p)=(1,1,2)}$ .

You don't need Zsigmondy as others have shown. But this lemma generalizes better. Or of course you can use Catalan's conjecture (now proved), which says (for $x,y,a,b>1$ ) $x^y-a^b=1\iff (x,y,a,b)=(3,2,2,3)$ .

So using it we have either $(x,y,p)=(2,2,3)$ or at least one of $p,x,y$ is $1$ .

$y=1$ gives $(x,y,p)=(1,1,2)$ and $x=1$ gives $y+1=p, y^2-y+1=1$ , the latter giving either $y=0$ , impossible, or $y=1$ , giving $(x,y,p)=(1,1,2)$ .

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Here is why Zsigmondy's Lemma generalizes better. Let's solve $p^x=a^n+b^n$ , where $p$ is prime, $x,a,b\ge 1, n\ge 2$ . Then either $(\{a,b\},n,p,x)=(\{2,1\},3,3,2)$ or $(a,b,n,p,x)=(2,2,n,2,n+1)$ or wlog $a>b>0$ , in which case Zsigmondy gives us a contradiction.

Using Catalan's Conjecture, we get the triplet (3,2,2): And one more i.e. (2,1,1)