# Thread: Solve a Integer equation system

1. ## Solve a Integer equation system

$a,b,c \in \mathbb{N}^+, \quad d = 2a^2 = 3b^3+2 = 5c^5+3$
Find the smallest $d$ above
What if $d = 2a^2+1 = 3b^3+2 = 5c^5+3$ ?

2. Originally Posted by elim
$a,b,c \in \mathbb{N}^+, \quad d = 2a^2+1 = 3b^3+2 = 5c^5+3$
Find the smallest $d$ above
If $2a^2+1 = 3b^2+2$ then $2a^2\equiv 1\!\!\!\pmod3$, which is impossible. So no solutions there. But the remaining equations $d = 2a^2+1 = 5c^5+3$ have at least one solution d = 163 (with a = 9 and c = 2).

3. Thanks Opalg! Let's forget that half of the problem
What about the smallest $d=2a^2=3b^3=5c^5+3$?

4. By Chinese Remainder Theorem,
$d = 38+30k$ for some $k$

5. Integer solution for $d=2a^2=3b^3+2=5c^5+3$
become very tough. I heard that $d < 10^{50}$ has no solutions

6. So $\frac{d}{2}$ is a square, $\frac{d-2}{3}$ is a cube, and $\frac{d-3}{5}$ is a fifth power.

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

- Hollywood