Question

advanced question:

Answer 1

uh huh

Answer 2

I Accept The Challenge -

Answer 3

Find the sum of all integers a such that \[a^2(2^a-a^3)+1\] is a power of 7

Answer 4

the answer is between 0 and 999 inclusive
I have it narrowed down to all possible values of a \[a = [94,46,96,58] (\mod100)\]
any ideas on what to do?

Answer 5

@bibby my savior help us.

Answer 6

Well - I Found That a = -1 - Etc....
http://www.wolframalpha.com/input/?i=a%5E2%282a%E2%88%92a%5E3%29%2B1 <--

Answer 7

Bibby's lost as I am.. :l

Answer 8

Just as i am*

Answer 9

I'm not lost this is easy-modo

Answer 10

Solve it step by step O.e

Answer 11

Would you be ok with a code solution?

Answer 12

well wait what is the question a2(2a−a3)+1 =a^7 right?

Answer 13

i have python, im more interested in the theory behind it, but yea that sounds cool @bibby

Answer 14

@thissucks its 7^x not x^7

Answer 15

@charlotte123 you plugged 2a not 2^a into wolfram

Answer 16

AH!

Answer 17

@Edutopia you wrote ' is a power of 7' which means a^7 are you sure it is 7^a

Answer 18

http://www.wolframalpha.com/input/?i=a%5E2%282%5Ea%E2%88%92a%5E3%29%2B1 <--- There You Go!

Answer 19

a power of 7 meaning 7^1 7^2 etc.

Answer 20

ohh I get it!

Answer 21

let me give it a try

Answer 22

power of 7 means 7^x

Answer 23

I know, from modular arithmetic, that the answers of a must have last 2 digits of 94, 46, 96, or 58 (besides a=1 which is 7^0)

Answer 24

any ideas?

Answer 25

First i noticed that a must be even, then i found that 7^x repeats {1,7,9,4} mod 10, then i found that it repeats 01,07,49,43 mod 100, (i did mod 1000 but it did not narrow down any further), so after having 01,07,49,43 i subtracted by one because of the equation and found that "a" when run through the function must be equivalent to 00,06,48 or 42 mod 100. i found the mod 10 sequences for 2,4,6,8 (because "a" must be even) and found that
"a" must be equivalent to either 94,46,96,or 58 mod 100 (when you plug these values into the function you get a number that is equivalent to either 00 or 48 mod 100)