Proving 2^(2^n) + 1 Ends in 7 for All n > 1 - Induction Question

[Kamataat]

Prove by induction that 2^{2^n}+1 always ends in 7 for all n > 1 (true for n = 2).

I couldn't figure out anything to do with the last digit being 7, so I looked the case that 2^{2^n} ends in 6 for all n > 1, which is also true for n = 2.

Suppose it's true for n = k:

2^{2^{k+1}}=2^{2^k\cdot 2}=(2^{2^k})^2

Now, since by assumption 2^{2^k} ends in 6, its 2nd power also ends in 6 since 6x6=36. Hence 2^{2^n}+1 ends in 7 for all n > 1.

Correct or not?

Thanks in advance,

- Kamataat

 

[HallsofIvy]

Yes, that's perfectly good. Obviously n+ 1 will end in 7 if and only if n ends in 6.

You could, just as easily, say "Suppose 2^{2^k}+1 ends in 7 for some k. Then 2^{2^k} ends in 6. Therefore, 2^{2^{k+1}}= (2^{2^k})^2 also ends in 6 and so 2^{2^{k+1}}+ 1 ends in 7".

 

[Benny]

I think what you've done is correct. Here is how I would set it out. I'm not sure if it's correct though.

Base case, assume true for n = so that 2^{2^k } \equiv 6\left( {\bmod 10} \right)

Using the hypothesis:

<br /> 2^{2^{\left( {k + 1} \right)} } \equiv 2^{2\left( {2^k } \right)} \equiv \left( {2^{2^k } } \right)^2 \equiv \left( 6 \right)^2 \equiv 6\left( {\bmod 10} \right)<br />

I think the result follows from that.

 

[DaveC426913]

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