Posted by robin! on Tuesday, February 26, 2013 at 11:36am.
How many pairs of integers (not necessarily positive) are there such that both a2+6b2 and b2+6a2 are both squares?

algebra  Navneet, Thursday, February 28, 2013 at 12:43pm
There exists only one solution and the whole challenge is in proving the fact that there does not exist any other solution. So here we go...
Let's add both these expressions. Then as per the problem,
7(a^2+b^2)=c^2+d^2
Clearly LHS is a multiple of 7 and so RHS must be a multiple of 7 in order for the solution to exist. Now, let's investigate the RHS.
Both c and d can be expressed as 7k, 7k+1, 7k+2,7k+3,7k+4, 7k+5,7k+6.
So, remainder when (c^2+d^2) is divided by 7 can be the same if one among the following is divided by7:
0, (1+2^2), (1+3^2), (1+4^2), (1+5^2), (1+6^2)
(2^2+3^2), (2^2+4^2), (2^2+5^2), (2^2+6^2)
(3^2+4^2), (3^2+5^2), (3^2+6^2)
(4^2+5^2), (4^2+6^2)
(5^2+6^2)
None of the above except 0 is divisible by 7.
Hence there exists only one solution for (a,b) > (0,0) 
algebra  Calvin Lin, Monday, March 4, 2013 at 1:19pm
This is a problem posted on Brilliant(dot)org
The above solution is wrong / incomplete.
Calvin Lin
Brilliant Challenge Master