Which pairs of positive integers (𝑥,𝑛) satisfy $𝑥^2−2^𝑛=153$?

My attempt: Rearrange to $x^2=2^n + 153$ and with $2^n\geq 2\$ it follows $x^2 \geq 155\$. The next square number is 169, so $x = 13$ and $n = 4$. A first solution. Since $2^n$ is even and 153 is odd, $x^2$ will be odd. So any candidate solution will have an even distance of $2m$ from a previous solution and the difference between these solutions is $(x+2m)^2 - x^2 = 4mx + 4m^2$. This difference can be expressed as a difference between two powers of 2, $4mx + 4m^2 = 2^p - 2^n$. My idea was to show that this doesn´t work so that the first solution is the only one.

• You accidentally wrote 155 a couple of times. – John Wayland Bales Sep 16 '18 at 23:12
• And $z$ for $x$.... Please be more careful in your postings. Otherwise you waste our time and lose credibility. – David G. Stork Sep 16 '18 at 23:15
• I apologize for the mistakes and re-edited the question. Sorry. – Parzifal Sep 16 '18 at 23:18
• The Ramanujan-Nagell equation is $x^2-2^n=-7$. Have a look at the literature on that equation, and see whether you can adapt it to your question. You can start with en.wikipedia.org/wiki/Ramanujan–Nagell_equation – Gerry Myerson Sep 16 '18 at 23:20
• You seem to switch from $2^n$ to $n^2$ a lot. Which one do you mean? – lulu Sep 16 '18 at 23:21

$n$ must be even.

$$153 = 3^2 \cdot 17$$

If $$x^2 - 2 y^2$$ is divisible by $3,$ then both $x,y$ are divisible by $3.$ Since this $y$ would be a power of $2$ this is impossible.

So $n$ is even, $n=2k,$ and we actually have $$x^2 - (2^k)^2 = 153 \; ,$$ $$(x+ 2^k) (x-2^k) = 153$$

Umm. $$(x+ 2^k) - (x-2^k) = 2^{k+1}$$

This leads to a finite set of possible $x,$ we can factor $153$ as (ordered pairs) $$153 \cdot 1$$ $$51 \cdot 3$$ $$17 \cdot 9$$

$$153 - 1 = 152 = 8 \cdot 19$$

$$51 - 3 = 48 = 16 \cdot 3$$

$$17 - 9 = 8$$

• Thank you - much appreciated. (It took me some time to figure out why n must be even – Parzifal Sep 17 '18 at 0:22

Suppose $x \geq 0$. Studying the equation modulo $2$ reveals that $x$ must be odd. Studying it modulo $3$ reveals that $n$ must be even $n=2k$.

$$x^2-2^n=(x-2^k)(x+2^k)=153$$

So $x-2^k$ and $x+2^k$ are two integers that average to $x$ but multiply to $153$. But $153=51(3)=1(153)=17(9)$, so either $x=26$ or $x=77$ or $x=13$. But $x$ is odd, so we must have $x=13$ or $x=77$. But $x=77$ doesn’t work.

Modulo 2: $x^2 \equiv 1$ so $x \equiv 1$.

Modulo $3$: $x^2-(-1)^n \equiv 0$ so $x^2 \equiv (-1)^n$. But since any square is equivalent to $0$ or $1$ modulo $3$. Either $(-1)^n \equiv 0$ (which is obviously impossible) or $(-1)^n \equiv 1$ (which is only possible if $n$ is even).

• Your explanation of why n is even was very helpful for understanding the solution - thank you. – Parzifal Sep 17 '18 at 0:27
• Glad to help! @Parzifal – Ahmed S. Attaalla Sep 17 '18 at 0:30