Question

Prove the following theorem. If n is a positive integer such that n ≡ 2 (mod 4) or n ≡ 3 (mod 4), then n is not a perfect square.

Answer 1

Suppose n is an positive integer of the form ( mod 4) .

n = 4k + 2 for some positive integer k .

Now if n is a perfect integer then there exist an positive integer m such that

. ............(i)

is even as it is a multiple of 2 .

for some positive integer a .

Substituting this values in (i) we get ,

  

a contradiction since left hand side is a multiple of 4 where right hand side is not so they cannot be equal .

So n is not a perfect square when (mod 4 ) .

.

Now suppose n is of the form ( mod 4 )

for some positive integer k .

Now if n is a perfect square then there exist a positive integer m such that

Now every positive integer can be written as one of the form 4k , 4k+1 , 4k+2 , 4k +3 . So m also can be written as one of these forms .

If m = 4k then

If m= 4k+1 then

If m = 4k +2 then

If m = 4k +3 then

So in each case cannot be written as of the form 4k +3 .

Hence , , a contradiction .

Hence n is not a perfect square if (mod 4) .