# Solve a Integer equation system

• May 4th 2010, 08:33 AM
elim
Solve a Integer equation system
$\displaystyle a,b,c \in \mathbb{N}^+, \quad d = 2a^2 = 3b^3+2 = 5c^5+3$
Find the smallest $\displaystyle d$ above
What if $\displaystyle d = 2a^2+1 = 3b^3+2 = 5c^5+3$ ?
• May 4th 2010, 12:49 PM
Opalg
Quote:

Originally Posted by elim
$\displaystyle a,b,c \in \mathbb{N}^+, \quad d = 2a^2+1 = 3b^3+2 = 5c^5+3$
Find the smallest $\displaystyle d$ above

If $\displaystyle 2a^2+1 = 3b^2+2$ then $\displaystyle 2a^2\equiv 1\!\!\!\pmod3$, which is impossible. So no solutions there. But the remaining equations $\displaystyle d = 2a^2+1 = 5c^5+3$ have at least one solution d = 163 (with a = 9 and c = 2).
• May 4th 2010, 04:09 PM
elim
Thanks Opalg! Let's forget that half of the problem
What about the smallest $\displaystyle d=2a^2=3b^3=5c^5+3$?
• May 4th 2010, 08:06 PM
elim
By Chinese Remainder Theorem,
$\displaystyle d = 38+30k$ for some $\displaystyle k$
• May 7th 2010, 04:52 PM
elim
Integer solution for $\displaystyle d=2a^2=3b^3+2=5c^5+3$
become very tough. I heard that $\displaystyle d < 10^{50}$ has no solutions
• May 8th 2010, 08:11 AM
hollywood
So $\displaystyle \frac{d}{2}$ is a square, $\displaystyle \frac{d-2}{3}$ is a cube, and $\displaystyle \frac{d-3}{5}$ is a fifth power.

If you look at these conditions modulo some small numbers, you might find something.

- Hollywood