Try this problem from Duke Math Meet 2009 Problem 7 based on the number of ordered triples. This problem was asked in the team round.
How many ordered triples of integers (a, b, c) are there such that 
and 
is divisible by 28.
Any square quantity is 0 or 1 modulo 4
Now the sum of three square quantities is 0 or 1 mod 4. Since 28 is 0 mod 4, therefore none of the three squares can be 1 mod 4.
Thus a, b, c are even.
Again a square quantity is always 0, 1, 4, 2 mod 7
Since we want the three squares to add up to 0 mod 7 (28 is 0 mod 7), the possibilities are (0, 0, 0) mod 7 and (1, 4, 2) mod 7
So we have the following cases:
Case 1
a, b, c are all even multiples of 7. There are 5 of them. Since we are looking for ordered triplet there are 
of these$
Case 2
a, b, c are all even numbers. with 1, 6 mod 7, 2, 5 mod 7 and 3, 4 mod 7
1 mod 7 - 10 elements
6 mod 7 - 10 elements
Hence there are 20 numbers which are 1 or 6 mod 7. 10 of these are even.
Similarly, there are 20 numbers which are 2 or 5 mod 7 and 20 numbers which are 3 or 4 mod 7. Half of each of which we will take.
So 10 choices for each set. 
choices. Now since we are taking ordered pairs, we must consider all the 3! = 6 permutations. So there are 6000 cases.
Hence answer is 6000 + 125 = 6125