Question

find factors of 17^100 + 32^201

Answer 1

is it 32 or 34?

Answer 2

32

Answer 3

i think the statement is incorect

Answer 4

vat do u think

Answer 5

me 2

Answer 6

so if there is 34 then , what would be the answer

Answer 7

we can factor out 17 then

Answer 8

how

Answer 9

17*2=34

Answer 10

\[17^{100} + 32^{201}\]\[=17^{(10^2)} + (2^5)^{201}\]

Answer 11

i have already done this but vat should be the next step

Answer 12

can you tell be where you found this question @singhmmm?

Answer 13

i have the self tutor book

Answer 14

what is the chapter/sub-chapter

Answer 15

polynomials

Answer 16

so what are the factors?

Answer 17

It can't be FACTORED :)

Answer 18

It should be 34 instead of 32 for FACTORING :)

Answer 19

The rough factorization though would be:
17^100 + 32^201

It is of the form
a^2 + 32b^4 where a= 17^50 and b=32^50

=> a^2 + 2(2b)^4
=>a^2 + 2c^2 where c = 4b^2 or c=2^2 * 2^500 = 2^502

now a^2 + 2c^2 = a^2 + ( (sqrt2)c )^2 + 2sqrt2 ac - 2sqrt2 ac
= (a+ sqrt2 c)^2 - [sqrt( 2sqrt2 ac)]^2
now again applying a^2-c^2 identity..
hmm..
long and tedious though
but hey, its factorized! :P

Answer 20

@shubhamsrg
But \(\sqrt{2}\) is not an integer, and usually it's implied that we only want integer factors in these types of problems. Finding real factors is sort of trivial.

Answer 21

hmm, what you say is right. I just couldn't find another legitimate solution.