Ques. for all

please check if the question is corrert?
wont the no. be 100497???
because the theory is any no. 1 more than multiple of 4 can be expressed as a sum of two integer's squares..............

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