Question

find all \(n\) such that \(17 | (8n^5-10)\)

Answer 1

@mathmath333 lets use below table and solve this
http://gyazo.com/444240ce9febfabbb76292d99d1a499a

Answer 2

http://www.wolframalpha.com/input/?i=primitive+roots+of+17

Answer 3

yes that table is using the primitive root 3
it has all indices for all the integers less than 13

Answer 4

this primitive roots stuff is above my level

Answer 5

thats okay, you're doing very good for an 1 hour of knowing them :)

Answer 6

lol

Answer 7

i feel you can work this problem with whatever you know so far

Answer 8

\[8n^5\equiv 10 \pmod{17}\]

Answer 9

take log both sides (ind)

Answer 10

\[\text{ind}~ (8n^5) \equiv \text{ind} ~10 \pmod{17}\]

Answer 11

use the properties and expand the keeping in mind that your goal is to solve "n"

Answer 12

\[\text{ind}~ (8n^5) \equiv \text{ind} ~10 \pmod{17}\]

use product property :

\[\text{ind}~ 8 + \text{ind}~n^5 \equiv \text{ind} ~10 \pmod{17}\]

Answer 13

next use exponent property

\[\text{ind}~ 8 + 5*\text{ind}~n \equiv \text{ind} ~10 \pmod{17}\]

Answer 14

from the table find out ind 8 and ind 10
plug them in above

Answer 15

u can write \(ind ~8=3ind ~2\) ?

Answer 16

ofcourse yes, but since we have the table with index values for all numbers, we can simply use them...

from the table ind 8 = 10, right ?

Answer 17

if u dont have a table then how can one solve them

Answer 18

we can create that table or we can solve it using the trick which you have found just now

Answer 19

ind 8 = 3*ind 2 = 3*14 = 42 = 10 (mod 16)

Answer 20

as you know creating that table is not that hard,
for now we may simply use that table..

Answer 21

it is coming \(\text{ind} ~n\equiv 2~mod~17\) ?

Answer 22

lets see

Answer 23

there is a typo earlier, let me fix it quick :

\[\text{ind}~ 8 + 5*\text{ind}~n \equiv \text{ind} ~10 \pmod{\color{Red}{16}}\]

Answer 24

now work it

Answer 25

how comes \( \color{Red}{16}\)

Answer 26

the indices properties work in mod\(p-1\)
16 = 17-1

Answer 27

ok i see

Answer 28

\[\text{ind}~ 8 + 5*\text{ind}~n \equiv \text{ind} ~10 \pmod{\color{Red}{16}}\]

from the table
ind 8 = 10
ind 10 = 3

plugging them in we have :
\[10+ 5*\text{ind}~n \equiv 3 \pmod{\color{Red}{16}}\]

Answer 29

subtract 10 both sides and see if you can solve "ind n"

Answer 30

if it helps replace "ind n" by \(x\) for time being :

\[10+ 5*x \equiv 3 \pmod{\color{Red}{16}}\]

solve \(x\)

Answer 31

\( 5*x \equiv 9 \pmod{\color{Red}{16}}\)

Answer 32

Yes! solve \(x\)

Answer 33

m stuck

Answer 34

find a value of \(x\) such that \(5x-9\) is divisible by \(16\)

Answer 35

wait \(x=21\)

Answer 36

we're in mod 16, so..

Answer 37

21 is same as 5 in mod 16
so x = 5 is the solution in mod 16

Answer 38

\[5*x \equiv 9 \pmod{\color{Red}{16}}\]

solving you get

\[x \equiv 5 \pmod{\color{Red}{16}}\]

but we let ind n = x earlier :

\[\text{ind} ~n = 5\]

Answer 39

look at the table and see which number has index as 5

Answer 40

5

Answer 41

Yes so the final solution is \[n \equiv 5\pmod{17}\]

Answer 42

that means all the integers of form \(5 + 17k\) where \(k\) is any integer, satisfy the given division problem

Answer 43

plugin n=5 and see if 8n^5-10 is really divisible by 17

Answer 44

ohk ,

Answer 45

yes it is

Answer 46

once you practice 1-2 problems, you will see why these work..

Answer 47

also this is a one line problem once you get familiar with the method

Answer 48

i can be familier with that only if i practice it by myself

Answer 49

seems like u r a ghost of ganeshie8

Answer 50

lol okay here few more problems for you to practice whenever you're free or want to
http://gyazo.com/07dd55c512d6a68a2097af4949bb8a8e

Answer 51

you can tag me or ganeshie anytime :)

Answer 52

ohk thnx :) for problems

Answer 53

ohk lol

Answer 54

you are right @mathmate333 ghost
i guess something wrong but dk

Answer 55

:'( cool problem