Ques. for all

well @Arnab -> Ya u r right .....!! well i've discussed bout this type of questions earlier....

I remember it now....

If some one wants the answer then they can read it in the hidden area below....

Any integer is of the form 2k or 2k+1
Now, there are three cases
Remaneder if
. (even)² + (odd)²
. (odd)² + (odd)²
. (even)² + (even)²
is divided by 4

so let's check out

case 1
(2k)² + (2k + 1)² = 4k² + 4k² + 4k + 1 = 8k² + 4k + 1
which on division by 4 leaves remainder 1

case 2
(2k + 1)² + (2k + 1)² = 2(4K² + 4K + 1) = 8K² + 8K + 2, which on division by 4 leaves remainder 2

case 3
(2k)² + (2k)² = 8k², which on division by 4 leaves remainder 0

So Conclusion, Sum of the squares of any two integers, on division by 4 leaves remainder 0,1,2

So le'ts check it on 100400, on division by 4 it leaves remainder = 0, so yes it can be written as the sum of two squares

Again, on 100497, on division by 4 it leaves remainder =1, so yes it can be written as the sum of two squares

And that's it....... Hope this was of some use to some of you

And that's it....... Hope this was of some use to some of you

Of course it is!