Determine all possible triplets of
nonzero perfect squares (A, B, C) that satisfy this equation:
1 2 3 2
 +  +  = 
A B C 3
Prove that there are no others.
(In reply to
re(2): Do i err ?? by Ady TZIDON)
None of A, B or C can be 1, since the LHS would be greater than 2/3.
For the same reason, C can't be 4.
When C = 9, we find a solution when A = B = 9.
You err in asserting that there's no need to check values bigger than 9. You haven't proven that there aren't other solutions where A = 4 or B = 4.

Posted by tomarken
on 20140529 09:41:25 