Question

Prove: If n ≡ 1 (mod 6), then n2 + 2n is composite, where n>1. @MIT 18.02 Multiva…

Answer 1

if
\[n\equiv 1( \text{ mod }6)\] then
\[n+2\equiv 3(\text{ mod } 6)\] and therefore
\[n(n+2)\equiv 3(\text { mod } 6)\] meaning it is divisible by 3

Answer 2

a more prosaic way of thinking about this is if
\[n\equiv 1 ( \text { mod } 6)\] then
\[n=1+6k\] so
\[n+2=3+6k\] making
\[n(n+2) = (1+6k)(3+6k)=3+24k+36k^2\] which is clearly divisible by 3

Answer 3

\[n^2+2n=n(n+2)\]
since this is factorable and is an integer
isn't n(n+1) composite for all integers >1

Answer 4

oops n(n+2)

Answer 5

good point. now that i think about it the problem is rather silly

Answer 6

you don't need the first part to prove it
its already done

Answer 7

just kidding

Answer 8

first type
i was just a answer checker for first problem
so there

Answer 9

first type of typo?

Answer 10

lol

Answer 11

of or or?

Answer 12

omg. I just realized part of the question was typed wrong! Sry! it is supposed to be n^2 + 2^n :(

Answer 13

i would use induction

Answer 14

\[\text{ so we have } n=6k+1\]
For k=1 we have n=8
\[8^2+2^8=64+256=320 \text{ this number is composite}\]
now assume it is true for some integer k=j we hae n=6j+1
this means this is true
\[(6j+1)^2+2^{6j+1}=(w)(r) \text{ for some integers w and r}\]
now we want to show it is true for k=j+1 we have n=6(j+1)+1
so
\[(6(j+1)+1)^2+2^{6(j+1)+1}=(6j+1+6)^2+2^{6j+1+6}\]
\[(6j+1)^2+2(6j+1)(6)+(6^2)+2^{6j+1} \cdot 2^6\]
\[(6j+1)^2+2^{6j+1} \cdot 2^6 +12(6j+1)+36\]
....
thinking...

Answer 15

i give up i can't do math anymore
i can't even evaluate n for k=1

Answer 16

which i know is n=7

Answer 17

i think we might have to use that if p is prime then it is congruent to 1 or 5 mod 6

Answer 18

i have to go do some stuff

Answer 19

we know n^2 is congruent to 1 mod 6 because n is

Answer 20

so it comes down to what is 2^n congruent to mod 6

Answer 21

k , thanks myininaya

Answer 22

the choices are only 2 and 4 because it is a power of 2

Answer 23

so maybe we can show that it is not 4, in which case we are done

Answer 24

ok

Answer 25

ah yes and it is not 4, it is 2

Answer 26

reason being that
\[2^{1+6k}=2\times 2^{6k}\] and
\[22^6=64\equiv 4(\text { mod } 6)\] and so (pretty sure of this but you should check
\[2^{6k}\equiv 4 (\text { mod } 6)\] so
\[2^{6k+1}\equiv 2( \text { mod } 6)\]

Answer 27

sorry i meant
\[2^6=64\equiv 4(\text{ mod } 6)\]

Answer 28

ok. So then the sum of n^2 + 2^n , which I think would be n^2 + 2^(6k+1), is composite?

Answer 29

i'm sry. I'm a little confused. :/

Answer 30

yes it is composite because a prime number must be congruent to 1 or 5 mod 6

Answer 31

n^2 is congruent to 1 mod 6 because n is

Answer 32

and
\[2^{1+6k}\] is congruent to 2 mod 6
so their sum is congruent to 3 mod 6 unless i made some mistake