**Argument**

*: Let a , b any integer, and z = a^2 - b^2*

*Then (z mod 4) can never be 2 for any value of a and b.***Proof**

Lemma 1.

*(A perfect square mod 4 ) is always either 0 or 1.*Proof for lemma 1.

Let k = any number.

Then 2k is an ever number, and (2k+1) is an odd number.

Square of 2k = 4k^2 ==> (4k^2) mod 4 = 0

Square of (2k+1) = (4k^2 + 4k + 1) ==> (4k^2 + 4k + 1) mod 4 = 1

*From lemma 1, we can conclude that (a^2 - b^2) can one of following four cases*

*1) even^2 - even^2*

*2) even^2 - odd^2*

*3) odd^2 - even^2*

*4) odd^2 - odd^2*

*where even = an even number and odd = an odd number.*Taking mod 4 for each case,

1) (0 - 0) mod 4 => 0

2) (0 - 1) mod 4 => 3

3) (1 - 0) mod 4 => 1

4) (1 - 1) mod 4 => 0

Therefore, for all possible cases, taking mod 4 of difference of squares always result in 0,1, or 3, but never 2.

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