[Guest]

Determine all possible pair(s) (p,q) of integers that satisfy this equation:

2p³ = 3q² + 4

Edited May 9, 2009 by K Sengupta

[Guest]

Determine all possible pair(s) (p,q) of integers that satisfy this equation:

2p³ = 3q² + 4

(p,q) = (2,2),(2,-2)

[Guest]

(p,q) = (2,2),(2,-2)

At this point, I am not saying whether these constitute the only possible pairs, since this would require a proof to validate that.

Spoiler for hint:

The proof is so simple that it is more of a “brainteaser” than a complicated mathematical exercise.

Edited May 11, 2009 by K Sengupta

[Guest]

I inserted this into my calculator by solving for q so everything is on the other side of q= and the lowest pair i got was (54,324)

[Guest]

I inserted this into my calculator by solving for q so everything is on the other side of q= and the lowest pair i got was (54,324)

Check your calculations.

2p³ = 3q² + 4

-> q² = (2p³ – 4)/3

For (p,q) = (54, 324),

lhs= 324² = 104976

rhs = (2*54³ – 4)/3 = 104974.66666666666………..,

which is a contradiction.

Edited May 11, 2009 by K Sengupta

[Guest]

You said the proof was simple but I just can't figure this one out. The only thing I can deduce is that both p and q must be even. Am I on the right lines or is it something staring me in the face?

[Guest]

You said the proof was simple but I just can't figure this one out. The only thing I can deduce is that both p and q must be even. Am I on the right lines or is it something staring me in the face?

Yes, the proof is simple.

It only involves simple manipulations and an elementary application of Fermat's Last Theorem.

[Guest]

Yes, the proof is simple.

It only involves simple manipulations and an elementary application of Fermat's Last Theorem.

Well that was actually the first thing I thought of but I couldn't see it going anywhere. I'll see what I can do.