Prove that if a positive integer n is a perfect square then

Prove that if a positive integer n is a perfect square, then n cannot be written in the form 4k + 3 for k an integer. Prove that no integer in the sequence 11.111.1111.11111.111111,... is a perfect square.

Solution

Case 1. m=4k for some integer k

m^2=16k^2=0 modulo 4

Case 2. m=4k+1, for some integer k

m^2=(4k+1)^2=16k^2+8k+1=1 modulo 4

Case 3. m=4k+2 for some integer k

m^2=16k^2+4+16k=0 modulo 4

Case 4. m=4k+3 for some integer

m^2=16k^2+24k+9=1 modulo 4

Hence all squares are either 0 or 1 modulo 4

So a perfect square cannot be 3 modulo 4 ie of the form 4k+3

(b)

To check divisibiliy by 4 we need only check the last two digits of a number. This can be easily seen

....n2n1n0=....+100*n2+10n1+n0

4|100 hence we need only check if: 10n1+n0 is divisible by 4

Hence,

111.....111 =11 modulo 4=3 modulo 4

But as proved in part (a) no perfect square can be 3 modulo 4.

Hence no integer in this sequence is a perfect square.