21please 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..............
Prove that 100400 can be written as the sum of two squares of integers....!!
Such an easy one....... but a good one though.......!!
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3 Answers
21
36well @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
71And that's it....... Hope this was of some use to some of you
Of course it is!