Find all possible triplets (A,B,C) of positive integers that satisfy the following system of equations:
A2 = 2(B+C) and:
A6 = B6 + C6 + 31(B2 +C2)
If you cube the first equation you can substitute
A6 in the second:
8(B+C)3 = B6 + C6 + 31(B2 +C2)
When I entered this into Wolfram|Alpha the graph is a rather small egg shape with the blunt end at b=1, c=1 and narrow end around b= 2.75, c=2.75
There are solutions very near (2,1) and (1,2) but not quite.
The only positive integer solution is (A,B,C)=(2,1,1).
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Posted by Jer
on 2015-06-05 11:08:59 |
Answer