x and y are a pair of positive integers such that both their sum and d

Let's start with value of positive squares below 100,

1,4,9,16,25,36,64,81..

given condition,
Sum and difference should be perfect square too.

Now observe that, for odd square there should be one even integer and one odd integer, (such that their sum or difference will be odd)

So every odd square will have sum or difference as odd. Similarly, even squares will have even integers or both odd integers such that sum/diff will be even.

Thus,

Starting with 1 = 0 pairs as there is no odd square less than 0 which can be the difference b/w two integers.

4 = again 0

9 = 1 pair as there is 1 odd square less than 9

Eg ( 5,4) = 5+4 = 9/5-4=1

16 = 1 pair as there is 1 even square less than 16

25 = 2 pairs (9,1)
Eg (13,12) ; (17,8)

36 = 2 pairs (16,4)

49 = 3 pairs ( 25,9,1)

64 = 3 pairs (36,16,4)

81 = 4 pairs ( 49,25,9,1)

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