divisibility.?

Question

prove that n^2 + 4n + 11 is not divisible by 49 for any integer n.

Answer 1

n^2 +4n + 11 = n^2 + 2*2n +4 -4 + 11 = (n + 2)^2 + 7

Because 7 is a prime number his number n^2 +4n + 11 is divisible by 7 iff (n + 2)^2 is divisible by 7 iff (n + 2) is divisible by 7.

Now, if (n + 2) is divisible by 7 (n + 2)^2 is divisible by 49, so n^2 +4n + 11 = (n + 2)^2 + 7 is divisible by 49 with the remainder 7.

Answer 2

Suppose n²+4n+11 = 0(mod 49).
Then n²+4n = 38(mod 49).
Complete the square to get.
n²+4n+4 = 42(mod 49).
(n+2)² = 42(mod 49),
u²= 42(mod 49).
But 42 is a multiple of 7, so u
must be a multiple of 7.
But if u is a multiple of 7, u² = 0(mod 49), so
this is impossible.