Prove by induction that 7^n + 4^n + 1 is divisible by 6 for n any positive integer?

First prove true for n = 1 which is trivial.

You now have to arrange 7^(n + 1) + 4^(n + 1) + 1 into something involving the original statement. the only idea that immediately comes to mind is

7^(n + 1) + 4^(n + 1) + 1 = 7*7^n + 4*4^n + 1

= 4*7^n + 4*4^n + 4 + 3*7^n - 3

= 4(7^n + 4^n + 1) + 3(7^n - 1)

Can you see how to proceed from here?

The last part must be a multiple of 3 and also a multiple of 2 (why?) so a multiple of 6.

I will start you off.

Let Pn be the statement: 7^n + 4^n + 1 is divisible by 6 for all positive integers n

To do this proof we need to do two things.

Basis Step 1. Prove P1 is true.

Inductive Step 2. Prove P(k+1) is true assuming Pk is true

P1 is easily checked to be true - just plug in 1 for n.

The inductive step is always the problem.,

We want to prove 7^(n+1) + 4^(n+1) + 1 is divisible by 6.

The trick is use the fact that 7^n + 4^n + 1 is dvisible by 6.

So you want to rewrite 7^(n+1) + 4^(n+1) + 1 = 7*7^n +4*4^n +1 =

Now note that if you add 6 to this number, you don't change whether or not it is divisible by 6.

Also if we add 3*4^n (which is divisible by 6), we don't change the divisibility,.

After those two tricks, I think you will see how to finish the proof.

work it out.

assertion is genuine for n = a million by way of indisputable fact that a million(a million+5)=6 is divisible via 6 assume the assertion is genuine for n = ok, then ok(ok^2 +5 )= 6m ,m an integer We now would desire to coach that (ok+a million)((ok+a million)^2 +5) is divisible via 6 (ok+a million)((ok+a million)^2 +5) = (ok +a million) (ok^2 + 2k +6) = ok(ok^2 +5 +a million + 2k) + ok^2 + 2k +6 = ok(ok^2 +5) +ok + 2k^2+ok^2 + 2k +6 = 6m+3k^2 + 3k + 6 = 6(m + a million) +3k(ok+a million) .........(a million) Now the two ok or (ok+a million) is even nonetheless 3k(ok+a million) is divisible via 6 so the expression in (a million) is divisible via 6 So (ok+a million)((ok+a million)^2 +5) is divisible via 6 and the assertion is genuine for n = ok + a million. consequently via induction n(n^2 +5) is divisible via 6 for all powerful integers n.