11

The problem goes as follows:

Find all possible pairs of $x,y,z \in \mathbb{N}$ which satisfy the equation $7^x+1=3^y+5^z$

My first instinct was to continue by modding, but I don't think I can get anything out of it. The obvious solutions seem to be $x,y,z=1$ and $x,y,z=0$, but I am not really sure how to approach the problem.

Thanks in advance!

Mike
  • 20,434
Peter Allen
  • 895
  • Computer experiment shows that those are the only solutions with $0\leq x,y,z < 100$ – saulspatz Jan 30 '21 at 19:56
  • 3
    @PeterAllen where did you get the problem? I know of at least one similar problem in a published book, where it turned out the authors did not know how to solve it. Of course, they thought they could do it when they first published – Will Jagy Jan 30 '21 at 21:47
  • 1
    By some considerations modulo 6,8,36 one gets $x,y,z$ odd (for values greater than 2) and $z-x \equiv 2 \pmod{6}$. Also, there are bounds $z \le \log_5(7) x, y \le \log_3(7) x, x \le \log_7(5) (x+y) $. Don't know if it can help. I hoped in an approach with increasing powers of 6 but from $6^3$ on there is a quadratic dependence on $x,z$ which is really unpleasant. – Andrea Marino Jan 31 '21 at 00:20
  • 1
    @Andrea Marino I was thinking something similar: that the bounds $y+z \le 4x$ would have to be used. It seems to be that the weaker condition $5^z +3^y -1 \mod 7^x =0$ has solutions in $y$ and $z$ no matter how large $x$ is. – Mike Jan 31 '21 at 01:16
  • Since 3 is a generator modulo 7 and 49, it is a generator modulo all powers of seven. The same holds for 5. Thus the weaker condition is satisfied and likely to be satisfied also for not-so-big $y$ and $z$, but I don't see how this prevents other modular obstructions to arise :'( P.S. To calculate the orders I used the website http://www.numbertheory.org/php/order.html – Andrea Marino Jan 31 '21 at 01:51
  • Sorry, maybe you meant a modular solution to exist with $y+z \le 4x$! In this case I am not sure that it exist, since $4x << \varphi(7^x)$, and also because the powers do not overcome $7^x$ (say a clock tour) many times with that constraint. Of course, this is very close to the original question.. – Andrea Marino Jan 31 '21 at 01:57

1 Answers1

3

The two solutions you found are indeed all of them. In fact, we can solve this with a straight mod bash, which was surprising to me.

If any of $x,y,z$ are $0$, then we easily get $x=y=z=0$. If $y=1$, then $7^x=5^z+2$, and we readily get that $z=1$, so $x=1$ too. So now assume $y>1$ and $x,z>0$. Let the mod bash begin.

Taking mod $3$ implies that $z$ is odd. Taking mod $4$ implies that $x$ and $y$ have the same parity. Taking mod $5$ then implies that $x\equiv y\equiv 1\pmod 4$.

Now taking mod $7$, we get that either (a) $y\equiv1\pmod{12}$ and $z\equiv1\pmod{6}$ or (b) $y\equiv5\pmod{12}$ and $z\equiv5\pmod6$.

Now taking mod $9$, in case (a) we have $x\equiv5\pmod{12}$ and in case (b) we have $x\equiv9\pmod{12}$.

Consider case (a). If $z\equiv1\pmod{12}$, then taking mod $13$ gives $11+1\equiv3+5\pmod{13}$; if $z\equiv7\pmod{12}$ we still get a contradiction $11+1\equiv8+5\pmod{13}$.

Similarly in case (b), we get a contradiction in mod $13$, considering both cases $z\equiv 5\pmod{12}$ and $z\equiv11\pmod{12}$.

jlammy
  • 9,164