Prove that n^2 + 3n + 5 is not divisible by 121.

Hi Aritra,

Let us assume that n²+3n+5 is divisible by 121 for some value of n which is an integer.

So n²+3n+5 must be divisible by 11.

Hence [n²-8n+16 + 11n-11] must be divisible by 11

Or (n-4)² + 11(n-1) must be divisible by 11.

As 11(n-1) is divisible by 11, then (n-4)² must be divisible by 11.

Which is possible only when n-4 is divisible by 11.

Or n-4 = 11x --------(for some integer x)

or n = 4+11x

For this n, n²+3n+5 = 121x²+121x+33 = 121x(x+1) + 33.

Which gives a remainder of 33, when divided by 121. And hence cannot be divisible by 121.

And hence our assumption is incorrect.

Which means n²+3n+5, can never be divisible by 121.

Hope it helps.

Best Regards,

Ashwin (IIT Madras).

Approved