Question

Prove: If n ≡1 (mod 6), then n^2 + 2^n is composite. @MIT 18.02 Multiva…

Answer 1

Given:
n ≡1 (mod 6)

Prove: n^2+2^n is a composite number.

Since: n is defined, and is 1 (mod 6):
[1 (mod 6)]^2+2^[1 (mod 6)]= 3

And since 3 is prime:

Lemma 1 fails via counter-proof.
QED

Answer 2

oh, i see, so the proposition was false?

Answer 3

It certainly appears so

Answer 4

lol! Ok, thanks! :)

Answer 5

1 (mod 6) =1
so,

1^2=1
2^1=2
1+2=3
3 is prime, not composite.

Answer 6

\[n=6m+1\quad,\quad m=1,2,3,\dots\]\[n\equiv 1(mod \,\,6)\quad\Rightarrow\quad n^2\equiv 1(mod \,\,6)\]\[n\equiv 1(mod \,\,6)\quad\Rightarrow\quad 2^n\equiv 2(mod \,\,6)\quad,\quad n-\mbox{odd number}\]\[\Rightarrow\quad n^2+2^n\equiv 3(mod \,\,6)\quad\Rightarrow\quad n^2+2^n=6l+3=3(2l+1)\]\[\Rightarrow\quad n^2+2^n\equiv 0(mod \,\,3)\]