(I) gives (x + y) = {1,11,21,...} or {6,16,26,...}

(II) gives (x - y) = {2,12,22,..} or {7,17,27,...}

(x^2 + y^2) = (x + y)^2 - 2xy = (x - y)^2 + 2xy

As we do not get any information about xy from either (I) or (II) we can rule out A and B.

Now combine them.

(x+y)^2(mod5) = 1 (as the last digit would be 1 or 6)

(x-y)^2(mod5) = 4 (as the last digit would be 4 or 9)

Add them and you would get 2(x^2 + y^2) on the LHS.

On the RHS you would get 1+4=5. So 2(x^2 + y^2)(mod 5) = 0 ==> (x^2 + y^2)(mod 5) = 0. So ans is

C.

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Thanks,

Zooroopa