Question

Prove that $n^2 + 3n + 5$ is never divisible by 121 for any positive integer ${n}$.

Answer 1

 Let us say there were indeed a whole number k such that n² + 3n + 5 = 121k. Then n² + 3n + (5 - 121k) = 0. 

Then by the quadratic formula, n = [-3 +/- √(9 + 484k - 20)] / 2. 

So the proof of the proposition amounts to proving that the discriminant is never a perfect square. That is, 484k - 11 is never a perfect square. 

Note that of course, 484 = 22², i.e. 22²k - 11 is never a perfect square. 

I think this is clear because if it were, then it would have to be the square of an odd multiple of 11, which means it would have to be a product of 121 and another odd number. 

Which means that (484k - 11)/121 would have to be a whole (odd) number. But this is impossible because this is 44k - 1/11, where 44k is whole, therefore the difference is not. Thus we obtain a contradiction, by which we have shown that the discriminant cannot be a perfect square, therefore there is no whole-number solution of n satisfying the initial proposition.

There's an alternate method too:

The given function: f(n) = n^2 + 3n + 5 
If divisible by 121 (assume) then: 
f(121) = (121)^2+3(121) + 5 should be equal to 0 but no, its equals to 15009 
hence, its is not divisible by 121 
AND also, RHS =121 is a positive value and LHS is also a positive value, hence another reason it can never be divisible by 121

Or,

f(n) = n² + 3n + 5 
f(121) = 121² + 3(121) + 5 
f(121) = 14641 + 363 + 5 
f(121) = 15009 

f(n) is divisible by n if and only if f(n) mod n = 0. Since 15009 is not divisible by 121, f(n) is never divisible by 121

Hope This Helps :)