Find all $x,y,z \in \mathbb{Z^{+}}$ such that $20^x+15^y=2015^z$

I was checking modulo $4$ to see that $y,z$ must have the same parity.

Then took two cases when both $y,z$ are even and when both are odd. But it is difficult to find solutions

  • $\begingroup$ There are no solutions for $x,y,z<200$ (using computer). Maybe doesn't exist solutions. $\endgroup$ – MonsieurGalois May 24 '16 at 9:34

$20^x+15^y=2015^z$ equiv. $4^x5^x+ 3^y 5^y=403^z 5^z$.

We get $x=y$ or $x=z$ or $y=z$ because of divisibility by 5 to a certain power. (Suppose, let's say, $x < y < z$ then $4^x =403^z 5^{z-x} - 3^y 5^{y-x}$ impossible because the left side is multiple of 5)

Case 1) $x=y$

This case $z \ge x$ (otherwise $(3^x + 4^x)5^{x-z}=403^z$, impossible because $403$ is not divisible by $5$)

and $3^x + 4^x=403^z5^{z-x} \tag1$

But $3^x + 4^x < 2\cdot4^x< 403^z$ so (1) cannot hold

Case 2) $y=z$

This case $x \ge y$ and $3^x5^{x-y} + 4^y=403^y \tag2$ From (2) $ 3^x5^{x-y}=403^y - 4^y=399 \cdot A = 3\cdot7\cdot19\cdot A \tag3$

So the equality (3) cannot hold.

Case 3) $x=z$

This case $y \ge x$ and $3^x + 4^y5^{y-x}=403^x \tag4$

Applying modulo $3$ to (4) we get $2^{y-x}=1 \mod 3$ so $y-x=2k, k \in \mathbb{N}$.

Also from (4) $ 4^y5^{y-x}=403^x - 3^x=400 \cdot A \tag5=4^25^2\cdot A$

So far, this case I couldn't get further.

  • $\begingroup$ Can you explain why $x=y$ or $x=z$ or $y=z$ must hold? $\endgroup$ – Legend Killer May 23 '16 at 19:03
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    $\begingroup$ @AyanBiswas $5^x (4^x-3^y 5^{y-x}) = 5^z 403^z$ $\endgroup$ – Narek Maloyan May 23 '16 at 19:07
  • $\begingroup$ @AyanBiswas See my updated answer $\endgroup$ – user261263 May 23 '16 at 19:07
  • $\begingroup$ @N.S.JOHN Do you mean $z \ge x$? $\endgroup$ – user261263 May 24 '16 at 8:10
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    $\begingroup$ From your (5), if $y-x > 2$, one has that $5 \mid x$ and so $51721 \mid 403^x-3^x$. It follows that $y-x=2$ which, after a little work, contradicts the original equation. $\endgroup$ – Mike Bennett May 24 '16 at 20:47

Using modulo 7, you have that

$$20^x+15^y\equiv_7 (-1)^x+1^y$$

But $2015^z\equiv_7 (-1)^z$, so you need to solve an equation that solves:

$$(-1)^x+1^y\equiv (-1)^z$$

And is easy to see that this doesn't have solutions: The only numbers you can have are:

$$1+1= 2\not\equiv \pm 1$$

$$-1+1= 0\not\equiv \pm 1$$

  • $\begingroup$ With this you have that there are not solutions for any three numbers $(x,y,z)$ $\endgroup$ – MonsieurGalois Jul 31 '16 at 4:53

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