How can I solve the equation $$y^{3}-3^{x}=100$$ over positive integers?
On natural solutions of the equation $y^{3}-3^{x}=100$
7
$\begingroup$
number-theory
elementary-number-theory
diophantine-equations
-
2I have some idea how to prove that no other solutions exist but it's clear that there is a solution $y=7$, $x=5$ because $243+100=343$. – 2011-06-05
-
1Yes, but how can I show that this solution is unique? – 2011-06-05
-
1This is a fun problem. We know that 3 must be the only prime divisor of $y^3-100$. I tried to compute the 3-adic expansion of $y$. At the moment I can show that if $y>7$, then $y$ must be congruent to $7+38\cdot 81$ modulo $3^9$. Effectively I've been calculating the digits of the cube root of 100 in the 3-adic ring $Z_3$. Another idea is needed. – 2011-06-05
-
0Huh... I was thinking of an easy solution... – 2011-06-05
-
1Guess and check. But for some reason I don't think that's the sort of solution method you want here. – 2011-06-05
-
0I don't have a solution, but I have a method that is giving me more and more information about $x$ the more I follow it. For example, I can show that $x$ is 5 mod 42, and that after $x=5$, no values of $x$ are possible solutions until $x=215$. Would you mind saying a little more about how/where this question arose? – 2011-06-11
-
0This problem just popped to my head and I just wrote it on my paper sheet to solve it later (because I was sure it doesn't have nice solutions). And after a while that I tried it, I saw the only solution is $(7,5)$ and I planned to ask it here. – 2011-06-13