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 sum of three square quantities are 0 or 1 mod 4. Since 28 is 0 mod 4, there fore 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 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
