Melody

Hayley, the last time I looked, that thing wasn't working. :)

Hayley, that is an excellent answer :D

I don't know - neither of those would work without modification. 

Maybe another mathematician can help you :/    Sorry.

Do you mean    pedmas ?

Your seconds question makes no sense to me.       :(

Another way to write this is 

\(x=(x^2-y)^{2016!} \)

x and y must both be integers

If x=0 and y=0    then 0=0  that works

If   \(x=1\;\;and\;\;x^2-y=\pm1\)   

that would work to if they are both integers that is

1-y=1

y=0

so

if  x=1 and y=0  that works too

1-y=-1

2=y

so

if x=1 and y=2  that will work

and that is it     

so the 3 answers are   (0,0)   (1,0)  and  (1,2)         

So x+y  = 0 or 1 or 3

Maybe another mathematician can do it a different / better way  ??

You might be screwing up order of operation :/

You find the area of the 6 triangle sides and the area of the hexagonal base and add them all together :)

Hello Hayley :)

If you add one to a prime then it will definitely NOT be prime. Unless the prime number is 2.

This is becasue 2 is the only EVEN prim number.

All other prime numbers are odd.

so

if you add 1 to any of them you will have an even number and the only even number that is prime is 2!

You may find this interesting :)

http://www.ted.com/talks/adam_spencer_why_i_fell_in_love_with_monster_prime_numbers