algebra
posted by robin! .
How many pairs of integers (not necessarily positive) are there such that both a2+6b2 and b2+6a2 are both squares?

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) 
This is a problem posted on Brilliant(dot)org
The above solution is wrong / incomplete.
Calvin Lin
Brilliant Challenge Master