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# FP2 Induction Watch

1. Prove that 7^n + 4^n + 1 is divisible by 6 for all positive integers n.

I showed it was true for n=1
Assumed true for n=k
7^k + 4^k + 1 = 6m where m is a positive integer
Now to prove that the statement is true for n=k+1
7^(k+1) + 4^(k+1) + 1
7x7^k + 4x4^k + 1
7x7^k +4(6m - 7^k -1) + 1
7x7^k + 24m -4x7^k -4 +1
3x7^k + 24m -3
Now this is the part I always get stuck on. How do I proceed to take a factor of 6 out now?
2. (Original post by thers)
Prove that 7^n + 4^n + 1 is divisible by 6 for all positive integers n.

I showed it was true for n=1
Assumed true for n=k
7^k + 4^k + 1 = 6m where m is a positive integer
Now to prove that the statement is true for n=k+1
7^(k+1) + 4^(k+1) + 1
7x7^k + 4x4^k + 1
7x7^k +4(6m - 7^k -1) + 1
7x7^k + 24m -4x7^k -4 +1
3x7^k + 24m -3
Now this is the part I always get stuck on. How do I proceed to take a factor of 6 out now?
Consider the first and third terms together and notice that .

Alternatively, start again by considering the expression . If you can show this is divisible by 6, then you're done (why?).
3. I tried that f(k+1)-f(k) method beforehand but it comes out as 6x7^k + 3x4^k which I can't take a factor of 6 out.
4. (Original post by thers)
I tried that f(k+1)-f(k) method beforehand but it comes out as 6x7^k + 3x4^k which I can't take a factor of 6 out.
Can you see a factor of 2 in that second term? Does this solve your problem?
5. (Original post by Farhan.Hanif93)
Can you see a factor of 2 in that second term? Does this solve your problem?
Unfortunately not. All I see is that 4^k can be written as 2^2k.
6. (Original post by thers)
Unfortunately not. All I see is that 4^k can be written as 2^2k.
Which can further be written as .
7. (Original post by Farhan.Hanif93)
Which can further be written as .
Ah right, got it now. Thank you.

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