Mathematical induction problem

Siya

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Hello!
I am really not sure where to post this problem, so here we go.

Prove by mathematical induction:

\(\displaystyle P(n)\, \equiv \, \left(3 \cdot 5^{2n+1}\right)\, +\, 2^{3n+1}\, \Rightarrow \, P(n) \vdots 17\)

I have been trying to solve this problem and, at this moment, I can say I have no idea what to do in order to prove above equation is always divisible by 17. I tried, I really did, but I am simply stuck. If someone could help me out, at least a little, I would be much appreciated!

Edit:
This is my attempt:
P(1) is correct.
P(k) assume correct.
Prove for P(k+1) =>
3 * 52(k+1)+1 + 23(k+1)+1 =
=3 * 52 * 52k+1 + 23k+1 * 23 = 75 * 52k+1 + 23k+1 * 8 =
=(17*4)52k+1 + 7*52k+1 + (7+1)23k+1 =
=(17*4)52k+1 + 7(52k+1 + 23k+1) + 23k+1
That would be the closest. Leaves me with extra 23k+1. What am I missing?
 
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Hello!
I am really not sure where to post this problem, so here we go.

Prove by mathematical induction:

\(\displaystyle P(n)\, \equiv \, \left(3 \cdot 5^{2n+1}\right)\, +\, 2^{3n+1}\, \Rightarrow \, P(n) \vdots 17\)

I have been trying to solve this problem and, at this moment, I can say I have no idea what to do in order to prove above equation is always divisible by 17. I tried, I really did, but I am simply stuck. If someone could help me out, at least a little, I would be much appreciated!

Try induction. It is true for n = 0
P(0) = 3 * 5 + 2 = 17
So assume true for n and show that implies it is true for n + 1.
P(n+1) = 3 * 52(n+1)+1 + 23(n+1)+1 = 52 3 * 52n+1 + 23 23n+1 = ...
 
Last edited by a moderator:
Try induction. It is true for n = 0
P(0) = 3 * 5 + 2 = 17
So assume true for n and show that implies it is true for n + 1.
P(n+1) = 3 * 52(n+1)+1 + 23(n+1)+1 = 52 3 * 52n+1 + 23 23n+1 = ...

That is where my problem begins, actually. When I get to that part, I am simply not sure what to do next. I tried representing 3, lets say, as (5-2), then 5 as 17-12, then adding etc. but leads me nowhere, really.
I was thinking to assume P(n+1) is not divisible by 17, P(n+1)/17=q+r, and to somehow prove it as a contradiction but I think that would be another dead end. :confused:

Thank you, tho. At least I know the very start is okay.
 
That is where my problem begins, actually. When I get to that part, I am simply not sure what to do next. I tried representing 3, lets say, as (5-2), then 5 as 17-12, then adding etc. but leads me nowhere, really.
I was thinking to assume P(n+1) is not divisible by 17, P(n+1)/17=q+r, and to somehow prove it as a contradiction but I think that would be another dead end. :confused:

Thank you, tho. At least I know the very start is okay.

P(n+1) = 3 * 52(n+1)+1 + 23(n+1)+1 = 52 3 * 52n+1 + 23 23n+1 = 25 * 3 * 52n+1 + 8 * 23n+1 = (8 + 17) * 3 * 52n+1 + 8 * 23n+1
and remember, the assumption is 17 divides P(n)
 
P(n+1) = 3 * 52(n+1)+1 + 23(n+1)+1 = 52 3 * 52n+1 + 23 23n+1 = 25 * 3 * 52n+1 + 8 * 23n+1 = (8 + 17) * 3 * 52n+1 + 8 * 23n+1
and remember, the assumption is 17 divides P(n)


Wth. Thank you so much. I never though of making 25=17+8...So ashamed right now. :oops: Better not look how I tried to solve it. Thx!!!
 
P(n+1) = 3 * 52(n+1)+1 + 23(n+1)+1 = 52 3 * 52n+1 + 23 23n+1 = 25 * 3 * 52n+1 + 8 * 23n+1 = (8 + 17) * 3 * 52n+1 + 8 * 23n+1
and remember, the assumption is 17 divides P(n)

That is brilliant....
 
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