Show that 4n2 1 is never divisible by 11 for any integer nS

Show that 4(n^2 + 1) is never divisible by 11 for any integer n.

Solution

Assume that 11 divides 4*(n^2 + 1) .

Since 11 is prime and 11 doesn\'t divide 4.

11 must divide (n^2 + 1).

Any integer can be written as n = 11q + r where q and r are integers and r = 0, 1, 2, 3, ...
n^2 + 1 = (11q + r)^2 + 1
= 121q^2 + 22qr + r^2 + 1
= 11*(11q^2 + 2qr) + r^2 + 1
Now since n^2+1 is divisible by 11 and obviously 11*(11q^2 + 2qr) is divisible by 11, r^2 + 1 must also be divisible by 11.

We can now run through all the cases for r:
0^2 + 1 = 1
1^2 + 1 = 2
2^2 + 1 = 5
3^2 + 1 = 10
4^2 + 1 = 17
5^2 + 1 = 26
6^2 + 1 = 37
7^2 + 1 = 50
8^2 + 1 = 65
9^2 + 1 = 82
10^2 + 1 = 101

None of these is divisible by 11, so r^2 + 1 cannot be divisible by 11, but this is a contradiction.

Hence proved