Need an opinion on this solution involving mathematical induction

[nistaria]

Hey guys, this is a problem given to us by our professor in one of the worksheets. I would like an opinion to see if this proof is valid

Homework Statement

Use mathematical induction to prove that 3$$^{n}$$+7$$^{n}$$-2 is divisible by 8 for $$\forall$$n$$\in$$$$Z$$$$^{+}$$

Homework Equations

Base step:
n=0
3⁰+7⁰-2=1+1-2=0
0 is divisible by 8 therefore it's true for n=0

The Attempt at a Solution

Inductive step
Assume that 3^k+7^k-2 is divisible by 8 is true.
We need to show that 3^k+1+7^k+1-2 is divisible by 8 as well

3^k+1+7^k+1-2 = 3*3^k+7*7^k-2
= 3(3^k+7^k-2)+4*7^k+4

By inductive hypothesis we already know that 3^k+7^k-2 is divisible by 8 .
Now this is my problem, I'm also trying to show that 4*7^k+4 is divisible by 8 as well. This is my attempt at trying to prove it

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

7^k+1 will always be an even number for k$$\ge$$0. Multiplied by 4 this number will always be divisible by 8.
My question is do I still have to prove that 4*(7^k+1) is divisible by 8 or is stating that is enough proof?

Thanks for reading

 

[statdad]

The setup looks good, but if I were grading this I'd look for more detail on why $$4 \left(7^k+1\right)$$ is divisible by 8.

I admit my choice is probably specific to my teaching style, but quite often I see students making a ``it is obvious that it is true'' claim when

a) it isn't true

or

b) it is true but they aren't sure why

I can tell you know why it's true - it shouldn't be that hard to write out.

(My choice for more work may, and probably will, be commented on as a little harsh by others: I admit it's a personal choice).

 

[nistaria]

Thanks for your input statdad, anything is appreciated! Knowing my teacher, she would look into specifics as well and that's why I had to ask.

I was thinking of proving that statement with another case by induction
( I gave up trying to use latex code, it's really messing up on me and not outputting the correct notations)

We are to prove that 4(7^k+1) is divisible by 8 for k is a positive integer

Base step
k=1
4(7^k+1)=4(7¹+1)=32
32 is divisible by 8

Inductive step
assume 4(7^j+1) for j=k and that it is true.
We need to show that 4(7^j+1+1) is true as well.
4(7^j+1+1)=4(7*7^j+1)
=28*7^j+4=4(7^j+1)+24*7^j
by the inductive hypothesis 4(7^j+1) is divisible by 8
24*7^j is also divisible by 8 as the former is a multiple of the latter.
Therefore 4(7^k+1) is divisible by 8

 

[statdad]

Thanks for your input statdad, anything is appreciated! Knowing my teacher, she would look into specifics as well and that's why I had to ask.

I was thinking of proving that statement with another case by induction
( I gave up trying to use latex code, it's really messing up on me and not outputting the correct notations)

We are to prove that 4(7^k+1) is divisible by 8 for k is a positive integer

Base step
k=1
4(7^k+1)=4(7¹+1)=32
32 is divisible by 8

Inductive step
assume 4(7^j+1) for j=k and that it is true.
We need to show that 4(7^j+1+1) is true as well.
4(7^j+1+1)=4(7*7^j+1)
=28*7^j+4=4(7^j+1)+24*7^j
by the inductive hypothesis 4(7^j+1) is divisible by 8
24*7^j is also divisible by 8 as the former is a multiple of the latter.
Therefore 4(7^k+1) is divisible by 8

More than I expected :)

I was thinking that if you could argue $$7^k + 1$$ is always even, then

$$7^k + 1 = 2m$$

for some integer m, so $$4\left(7^k + 1 \right) = 8m$$ is divisible by 8.

 

[nistaria]

More than I expected :)

I was thinking that if you could argue $$7^k + 1$$ is always even, then

$$7^k + 1 = 2m$$

for some integer m, so $$4\left(7^k + 1 \right) = 8m$$ is divisible by 8.

Oh I like that.. very simple and effective!
Thanks for the input, I am much obliged.

 

[hunt_mat]

You could state that the product of two or more odd numbers is itself odd, that should be enough.