Question

Fun!

Answer 1

Prove that \(\tau(p+1)\) is never a prime number except when p=2.

Answer 2

tau

Answer 3

is the first value `r`?

Answer 4

\[\tau(n)\] is the total number divisors of n, so:

\[\tau(12)=6\] since it has 6 divisors, 1, 2, 3, 4, 6, and 12.

Answer 5

Ah gotcha

Answer 6

if tau(p+1) is odd then p+1 is perfect square

Answer 7

oh lol

Answer 8

that looks nice

Answer 9

hey wait
if p=16
then
\[\tau(17)=2\]
divisors will be 1 and 17
so... p can be any number 1 less than prime

Answer 10

p+1 can't be a perfect square...

if x^2 = p+1 => x^2-1 = p => (x-1)(x+1) = p => p is not prime

Answer 11

ok we had this from last day