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

1. amistre64

sounds like some sort of induction process to me

2. anonymous

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

3. 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

4. amistre64

i think your induction is off

5. 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.

6. amistre64

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

7. anonymous

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

8. amistre64

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

9. anonymous

@AlexADB definitely got the right idea

10. amistre64

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

11. amistre64

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

12. anonymous

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

13. anonymous

take out the 25 might work

14. anonymous

$25\times 5^{2k}-32\times 2^{5k}$

15. anonymous

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

16. anonymous

seen it before is how i know

17. amistre64

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

18. 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

19. amistre64

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

20. 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

21. amistre64

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

22. amistre64

both *terms* are factorable by 7 that is

23. amistre64

does it make sense yet?

24. anonymous

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

25. 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)

26. amistre64

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

27. 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?

28. amistre64

your induction is off again ...

29. amistre64

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

30. 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?'

31. amistre64

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

32. amistre64

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

33. 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?

34. amistre64

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

35. 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)

36. 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?

37. 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.

38. 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(5^2k−2^5k)−7×2^5k i don't see it. If you have time that would be awesome.

39. 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

40. amistre64

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

41. zepdrix

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