Find all triples of positive integers (x,y,z) such that

Question

Find all triples of positive integers (x,y,z) such that

$x^{z+1} \ - \ y^{z+1}=2^{100}$

The RHS is even, then x and y must be odd and $x^{z+1}>y^{z+1}$, but how to find out them all ?

Answer 1

You can factor $x^{z+1}-y^{z+1}$ by dividing out $x-y$. This shows that $x-y$ is a power of $2.$ But they must only have the same parity-they can both be even. In particular, there are many solutions with $z=0$ If $z=1$ you can factor it as $(x-y)(x+y)=2^{100}$ and you can find a finite number of solutions here. I think, but have not proved, that you will not find any with larger $z$.

Answer 2

$x \ge 0 \land y \ge 0 \rightarrow x \ge y$.

$$x^{z+1} - y^{z+1} = \left(x - y\right)\left(x^z + x^{z-1}y + x^{z-2}y^2 + \dots +y^z\right) = 2^{k+1} \tag{T1}$$

So we can conclude $x - y = 2^{r_1}$, and that $x$ and $y$ have the same parity.

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If $z$ is even, rewrite $(T1)$:

$$x^z + y^z + xy\underbrace{\left(x^{z-2} + x^{z-3}y + \dots + y^{z-2}\right)}_{\text{An odd number of terms}} = 2^{k} \tag{T2}$$

Since $x$ and $y$ have the same parity then $x$ and $y$ must both be even to satisfy the parity of $(T2)$. Let $2a = x$ and $2b = y$.

$$a^{z+1} - b^{z+1} = 2^{k - z}$$

which is a smaller instance of the original problem with the added assumption that $z$ is even. We can repeat the proof to get either $\exists c\, d\, : 2c = a \land 2d = b$ or that $2^{k - \dots}$ is no longer an integer, ad nauseam.

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If $z$ is odd,

$$x^{z+1} - y^{z+1} = \left(x + y\right)\left(x^z - x^{z-1}y + x^{z-2}y^2 - \dots -y^z\right) = 2^{k}$$

Conclude odd $z$ leaves a system of equations $$x-y = 2^{r_1}$$ $$x+y = 2^{r_2}$$

$x = 2^{r_2 -1} + 2^{r_1 - 1}, y = 2^{r_2 -1} - 2^{r_1 - 1}$

Expanding $x^{z+1} - y^{z+1} = 2^{k+1}$ for $z=1$ gives a set of solutions: $$2^{r_1 + r_2} = 2^{k+1}$$ $$(x,y,z) \in (2^{n - 1} + 2^{k - n},2^{n - 1} - 2^{k - n},1)\, 1 \le n \le k$$

Expanding for $z = 3$ gives another set:

$$2^{3r_2 + r_1 - 1} + 2^{3r_1 + r_2 - 1} = 2^{k+1}$$

with $(x,y,z)$ being found by the $r_1$ and $r_2$ values that satisfy the above.

At this point I'm unsure how to continue. All further expansions $z > 3$ seem to involve only positive terms, some of the form $A\cdot 2^{B(r_1, r_2)}\, ,\, A \ne 2$, which comes from the binomial expansion. It might be provable by induction. If so, then there are no more solutions.