Hi,
can anybody tell me how can i prove this :
26^(6n+5)+2*47^(12n+2)+3 = 0 mod(7)

Question

Hi,
can anybody tell me how can i prove this :
26^(6n+5)+2*47^(12n+2)+3 = 0 mod(7)

anonymous · Answer

$$26^{6n+5}+2*47^{12n+2}+3 ≡ 0 \mod(7)$$

ganeshie8 · Answer

$\large   26^{6n+5}+2*47^{12n+2}+3 \equiv  26^5 * 26^{6n} + 2*47^2*47^{12n} + 3 $

$\large   \equiv  (-2)^5 * 1 +  2 * (-2)^2 * 1 + 3 \equiv  0 \mod  7  $

anonymous · Answer

there is a question before this it was asking for the reminders of devision(5^n,7)

anonymous · Answer

@ganeshie8 i did not understand :D

ganeshie8 · Answer

By  FLT :
$a^{p-1} \equiv   1 \mod   p$

$47^6  \equiv  1 \mod  7$

and 

$26^6  \equiv  1 \mod  7$

ganeshie8 · Answer

FLT = Fermat Little Theorem..

u heard of it before right ?

anonymous · Answer

those are ok, go ahead :)

ganeshie8 · Answer

we're done :)

ganeshie8 · Answer

go thru it again :)

anonymous · Answer

ok, i mean what is the relation between the 47^6n+5 and the one that you have already typed ?

ganeshie8 · Answer

from where u got 47^6n+5 ?

anonymous · Answer

47^(12n+2) sorry :)

ganeshie8 · Answer

okay, lets go bit slow,  step by step  :)

ganeshie8 · Answer

$\large   26^{6n+5}+2*47^{12n+2}+3 \equiv  26^5 * 26^{6n} + 2*47^2*47^{12n} + 3 $

ganeshie8 · Answer

thats the first step, fine so far ?

anonymous · Answer

good mod 7 don't forgot :)

ganeshie8 · Answer

yes, we're in mod 7 world oly... just saving the typing energy lol ;)

anonymous · Answer

heh go ahead

ganeshie8 · Answer

second step :
$26^5 * 26^{6n} + 2*47^2*47^{12n} + 3  \equiv 26^5 * (26^{6})^{n} + 2*47^2*(47^6)^{2n} + 3  $

still fine ?

anonymous · Answer

mod 7 lol very find :) go ahead , i'm afraid from the third one

ganeshie8 · Answer

in  the third step, we apply FLT, we will use below :
$26^6 \equiv 1 \mod 7$
$47^6 \equiv 1 \mod 7$

ganeshie8 · Answer

in the third step, the thing inisde parenthesis, ima replace wid 1.

ganeshie8 · Answer

third step :-

$\large 26^5 * (26^{6})^{n} + 2*47^2*(47^6)^{2n} + 3 \equiv  26^5 * (1)^{n} + 2*47^2*(1)^{2n} + 3 $

anonymous · Answer

FLT = a+b=c mod n , a=c mod n and b=c mod n ?

ganeshie8 · Answer

FLT = Fermat Little Theorem :
$a^{p-1} \equiv   1 \mod p$

anonymous · Answer

ah ok

ganeshie8 · Answer

u sure, u covered Fermat topic yet ?

ganeshie8 · Answer

if u dont want to use Fermat, we can probably avoid it, but the proof will become nastly

anonymous · Answer

we didn't i don't know why

ganeshie8 · Answer

oh, then we should not use Fermat. 
Lets try to prove it without using Fermat.. let me think a bit

anonymous · Answer

flt is not that difficult, i don't know why they don't teach us, even it's proof

ganeshie8 · Answer

which course you doing ?

anonymous · Answer

i have studied the z devisions and the congruence as they call it

ganeshie8 · Answer

okay, lets forget Fermat for now.
they will teach u sooner or later.

lets try and prove it without Fermat  :)

ganeshie8 · Answer

Our first two steps remain same :-

$\large   26^{6n+5}+2*47^{12n+2}+3 \equiv  26^5 * 26^{6n} + 2*47^2*47^{12n} + 3 $

$26^5 * 26^{6n} + 2*47^2*47^{12n} + 3  \equiv 26^5 * (26^{6})^{n} + 2*47^2*(47^6)^{2n} + 3  $

anonymous · Answer

ok, i think that there is a relation between the 5^n and 7 devision and this stuff

ganeshie8 · Answer

lets see

ganeshie8 · Answer

next, take $(26^6)^n \mod  7$

anonymous · Answer

$$5^{6n+a}≡5^a \mod 7$$

ganeshie8 · Answer

we wont be needing that for this proof

ganeshie8 · Answer

lets try to simplify $(26^6)^n \mod  7$

ganeshie8 · Answer

$(26^6)^n \mod  7$

is same as 

$((-2)^6)^n \mod  7$

ganeshie8 · Answer

fine ?

ganeshie8 · Answer

cuz 26 leaves a remainder of -2 when divided by 7

anonymous · Answer

a mod n means a=0 mod n ??

ganeshie8 · Answer

$ a = b  \mod  n  $

means

$n $ divides   $a-b$

ganeshie8 · Answer

$26 = -2 \mod  7 $

means  

$7$ divides   $26--2$

anonymous · Answer

i know, you have typed (26^6)^n mod 7 what does this mean ?

ganeshie8 · Answer

that means we're trying to find the remainder when $7$ divides  $(26^6)^n$

anonymous · Answer

aha sorry i didn't understand that :D go ahead you are ok

ganeshie8 · Answer

$(26^6)^n \mod  7$

is same as

$((-2)^6)^n \mod  7$

$(64)^n \mod  7$

ganeshie8 · Answer

whats the remainder when 64 is divided by 7 ?

anonymous · Answer

=-(int(64/7)*7)+64 lol

ganeshie8 · Answer

?? no programming ok

anonymous · Answer

no programming = 1

ganeshie8 · Answer

good :)

ganeshie8 · Answer

$(26^6)^n \mod  7$

is same as

$((-2)^6)^n \mod  7$

$(64)^n \mod  7$

$(1)^n \mod  7$

$1 \mod  7$

thus,  $(26^6)^n \equiv  1 \mod  7$

ganeshie8 · Answer

plug this in 3rd step

anonymous · Answer

mm so you can type 1 in the place :)

ganeshie8 · Answer

yess so our proof goes like this so far :-

$\large   26^{6n+5}+2*47^{12n+2}+3 \equiv  26^5 * 26^{6n} + 2*47^2*47^{12n} + 3 $

$\large   \equiv 26^5 * (26^{6})^{n} + 2*47^2*(47^6)^{2n} + 3  $

$\large   \equiv 26^5 * (1)^{n} + 2*47^2*(47^6)^{2n} + 3  $

ganeshie8 · Answer

make sure you're at peace wid the proof so far :)

anonymous · Answer

yeah sure, and the same skill to 47

ganeshie8 · Answer

yup :)  next, take $(47^6)^{2n} \mod  7$
and simplify

anonymous · Answer

1^n whatever n is you got 1 so 26^5 ....... etc = 0 mod 7 GOT IT

ganeshie8 · Answer

yess, but u havent simplified  yet : $(47^6)^{2n} \mod  7$

ganeshie8 · Answer

simplify and conclude the proof properly lol

ganeshie8 · Answer

if u say etc etc... .ur professor wont give u full marks :P

anonymous · Answer

heh etc = openstudy , full response = paper

anonymous · Answer

Do you have facebook or a google account ?

ganeshie8 · Answer

then fine, i was worried if ur going to give ur professor etc etc ;)

ganeshie8 · Answer

msg me your gmail, ill add u :)

anonymous · Answer

let me close the Q