Question

Prove that 4n² + 4 = 0 (mod 19) for any n. ​

Answer 1

Answer:

Prove that 4n2+4 is not divisible by 19 for every n∈N .

More generally, for primes q of the form 4k+3 and any n∈N , q∤(n2+1) .

Suppose, to the contrary, that q∣(n2+1) for some prime q of the form 4k+3 and some n∈N . Then q∤n , and we have

1≡nq−1=(n2)(q−1)/2≡(−1)(q−1)/2≡−1(modq) .

This impossibility proves the claim.

Answer 2

Answer:

the above one is the right answer of your question btw good afternoon have a nice day