Solve a Integer equation system

Mar 2010
91
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usa
\(\displaystyle a,b,c \in \mathbb{N}^+, \quad d = 2a^2 = 3b^3+2 = 5c^5+3\)
Find the smallest \(\displaystyle d\) above
What if \(\displaystyle d = 2a^2+1 = 3b^3+2 = 5c^5+3\) ?
 
Last edited:
Aug 2007
4,041
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Leeds, UK
\(\displaystyle a,b,c \in \mathbb{N}^+, \quad d = 2a^2+1 = 3b^3+2 = 5c^5+3\)
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).
 
Mar 2010
91
16
usa
Thanks Opalg! Let's forget that half of the problem
What about the smallest \(\displaystyle d=2a^2=3b^3=5c^5+3\)?
 
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Mar 2010
91
16
usa
By Chinese Remainder Theorem,
\(\displaystyle d = 38+30k\) for some \(\displaystyle k\)
 
Mar 2010
91
16
usa
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
 
Mar 2010
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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