Prove that for all integers n>0, 5^(2n)-2^(5n) is divisible by 7.

- anonymous

Prove that for all integers n>0, 5^(2n)-2^(5n) is divisible by 7.

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- amistre64

sounds like some sort of induction process to me

- anonymous

It is an induction problem. I am typing out the work I have down, now.

- anonymous

Base Case: 5^((2)(1))-2^((5)(1)) is divisible by 7.
5^(2)*5-2^(5)*2=61
61|7
Inductive Hypothesis: n=k
5^((2)(k+1))-2^((5)(k+1))
5^((2)(k)*5-2^((5)(k))*2

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## More answers

- amistre64

i think your induction is off

- anonymous

I am not sure how to proceed from here. I know that I need to get 5^((2)(k))-2^((5)(k) because we have assumed that is divisible. Then the rest needs to be proved | by 7.

- amistre64

5^(2(k+1)) - 2^(5(k+1))
5^(2k +2) - 2^(5k +5)
5^(2k) 5^2 - 2^(5k) 2^5

- anonymous

Okay, let me try to work that out. I'll post what I have in couple moments.

- amistre64

5^(2n) - 2^(5n)
[5^2]^n - [2^5]^n
this might be a better format to play with

- anonymous

@AlexADB definitely got the right idea

- amistre64

5^(2(k+1)) - 2^(5(k+1))
[5^2]^(k+1) -[2^5]^(k+1)

- amistre64

but yeah, ive got no clear direction for this in my head yet

- anonymous

well, that was annoying. i had a problem posting for a second. okay, still trying to figure this out.

- anonymous

take out the 25 might work

- anonymous

\[25\times 5^{2k}-32\times 2^{5k}\]

- anonymous

\[25(5^{2k}- 2^{5k})-7\times 2^{5k}\]

- anonymous

seen it before is how i know

- amistre64

i was trying to divide it by 5^k - 2^k to see where it led :)

- anonymous

i am trying to piece together the steps you guys are throwing around. i'm not good at math so i'm having an issue keeping up :X

- amistre64

32 2^(5k) = 25 2^(5k) +7 2^(5k)

- amistre64

25 5^(2k) - (32 2^(5k))
25 5^(2k) - (25 2^(5k) +7 2^(5k))
25 5^(2k) - 25 2^(5k) -7 2^(5k)
[25 5^(2k) - 25 2^(5k)] -7 2^(5k)
etc ...
it was not obvious to me at first either

- amistre64

both sides are factorable by 7, and therefore ...

- amistre64

both *terms* are factorable by 7 that is

- amistre64

does it make sense yet?

- anonymous

im copying that onto paper. i need another 2 minutes to look at it.

- amistre64

25x + 32y
32y = 25y + 7y
25x + 25y + 7y
25(x + y) + 7y
but it was assumed that 7|x+y
25*7s + 7y
7(25s + y)

- amistre64

not sure if thats the proper method, but its proof enough to me

- anonymous

Base Case: 5^((2)(1))-2^((5)(1)) is divisible by 7. 5^(2)*5-2^(5)*2=61 61|7
Inductive Hypothesis: n=k 5^((2)(k+1))-2^((5)(k+1)) 5^((2)(k)*5-2^((5)(k))*2
Okay, I kind of need to see this from the beginning. First, is my base case and inductive hypothesis right?

- amistre64

your induction is off again ...

- amistre64

might help to get rid of the 'bases' and use the integer representations

- anonymous

Which part of my induction is wrong? Is it the base case? Is it the Hypothesis? What do you mean by getting rid of the 'bases?'

- amistre64

for example ...
5^((2)(k+1)) is not 5^(2k) * 5

- amistre64

5^(2k) is a base of 5, to a power if 2k ...
5^(2k) = 25^k

- anonymous

i didn't know i couldn't do that. 5^2 is the same as 5x5. I just reach up and grab a 5 from the exponent when I need more stuff to make the equation work, no?

- amistre64

5^(2(k+1)) = 5^(2k+2) = 5^(2k) * 5^2
you are not distributing correctly and it is throwing off your setup

- mathmate

Alternatively, as you did it:
\(S(n)=5^{2n}-2^{5n}\)
=>
Base case: n=1
\( S(1)=5^{2\times 1}-2^{5\times 1}=25-32=-7\) => 7|S(1)

- anonymous

@mathmate
Thank you. I like your notation mathmate.
Next, my Inductive Hypothesis would be:
5^(2(k+1))-2^(5(k+1)) I don't think there is any trouble there.
Inductive Step:
5^(2k)*5-2^(5k)*2
After this, I am not quite sure what to do. I want 5^2k-2^5k because that assumption is being treated as true. But, what should be done with the 5 and 2 that are in the way?

- mathmate

A lot of your misery is in mis-distribution. Once that's fixed, you can go along with what @Amistre64 suggested.
For the Inductive Hypothesis, it's just for the general case "n", or
\( 7~|~(5^{2n}-2^{5n})\), or simply 7|S(n)
For the Induction step,
You have to show that 7|S(n+1), \(given\) that 7|S(n).
Re-read @Amistre64's last few posts for hints on this step.

- anonymous

@satellite73
hey man, I still don't understand what is going on in this problem.
I understand 5^2k * 5^2 - 2^5k * 2^5; so, doing 25 and 32 i understand. but, getting here 25(5^2kâˆ’2^5k)âˆ’7Ã—2^5k i don't see it. If you have time that would be awesome.

- amistre64

Lets just clean up the expression
\[\Huge 25(\underbrace{5^{2k}}_{\color{red}{=x}})-32(\underbrace{2^{5k}}_{\color{green}{=y}})\]
Now, we have 25x - 32y ... and we want to factor out (x-y)
we can just as easily add zero to this setup,say (7x-7x)
25x + (0) - 32y
25x + (7x-7x) - 32y
25x + 7x -7x - 32y
(25x + 7x) -7x - 32y
32x -7x - 32y
32x - 32y - 7x
(32x - 32y) - 7x
32(x - y) - 7x

- amistre64

it is assumed, or given, that (x-y) is divisible by 7 ... so it is some value 7s.

- zepdrix

waow that's some pretty LaTeX :) lol
underbrace? Hmm never seen that before :3

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