Question

Prove that `(5^(2n+1)+2^(2n+1))` is divisible by `7 \ \ AA n in NN`?

I've been working on this for about an hour now and just can't seem to come up with a solution. Any ideas?

Answer 1

You could use induction.

Explanation:

The proof is a little tricky, so I've typed something up below in case you would like a solution.

Proof. We will prove by induction that, `AA` `n>=0 in NN`,

`color(white)(line break)`

( 1 ) `color(white)(spacespacespace)7|(5^(2n+1)+2^(2n+1))`

`color(white)(line break)`

By the definition of "divides," `EE` `m in ZZ` s.t. `7m=5^(2k+1)+2^(2k+1)`.

Base case : When `n=1`, the right-hand side of ( 1 ) is `5^3+2^3=125+8=133`, and `7|133`. So, ( 1 ) is true for `n=1`.

Induction Step: Let `k in ZZ` be given and suppose ( 1 ) is true for `n=k`. Then

`7m=5^(2k+1)+2^(2k+1)`

`=>7m=5^(2(k+1)+1)+2^(2(k+1)+1)`

`=>7m=5^(2k+3)+2^(2k+3)`

`=>7m=5^2*5^(2k+1)+2^(2k+3)`

`=>7m=5^2(5^(2k+1)+2^(2k+1))-25(2^(2k+1))+4(2^(k+1))`

`=>7m=5^2(7m)-21(2^(2k+1))`

`=>7m=7(25m)-7(3*2^(2k+1))`

`=>7m=7(25m-3*2^(2k+1))`

By the closure property of integers, `(25m-3*2^(2k+1)) in ZZ`. Thus, ( 1 ) holds for `n=k+1` and the proof of the induction step is complete. `color(white)(spacespacespacesp)` ∎

Answer 2

Proof by Mathematical Induction

Explanation:

Define `U_n` by;

`U_n = 5^(2n+1) + 2^(2n+1)`

Then our aim is to show that `U_n` is divisible by `7 AA n in NN`

We can prove this assertion by Mathematical Induction

When `n=0` the given result gives:

`U_n = 5^1 + 2^1 = 7`

So the given result is true when `n=0`

Now, Let us assume that the given result is true when `n=k`, for some `k in NN`, in which case for this particular value of `k` we have:

`U_k = 5^(2k+1) + 2^(2k+1)`

And `U_k` is divisible by `7`, and so:

`5^(2k+1) + 2^(2k+1)=7m`, say for some `m in NN`

`=> 2^(2k+1) =7m-5^(2k+1)" " ..... (star)`

So, Using the definition of `U_n` we have:

`U_(k+1) = 5^(2(k+1)+1) + 2^(2(k+1)+1)`
`" " = 5^(2k+3) + 2^(2k+3)`
`" " = 5^(2k+3) + 2^(2k+1+2)`
`" " = 5^(2k+3) + 2^(2k+1)*2^2`
`" " = 5^(2k+3) + 4*2^(2k+1)`
`" " = 5^(2k+3) + 4*(7m-5^(2k+1)) " "` (using `star`)
`" " = 5^(2k+3) + 4*7m-4*5^(2k+1)`
`" " = 5^(2k+1+2) + 4*7m-4*5^(2k+1)`
`" " = 5^(2k+1)*5^2 + 4*7m-4*5^(2k+1)`
`" " = 25*5^(2k+1) + 4*7m-4*5^(2k+1)`
`" " = (25-4)*5^(2k+1) + 4*7m`
`" " = 21*5^(2k+1) + 4*7m`
`" " = 7*3*5^(2k+1) + 4*7m`
`" " = 7(3*5^(2k+1) + 4m)`

Which is also clearly divisible by `7`

So, we have shown that if `U_n` is divisible by `7` for `n=k in NN`, then `U_n` is also divisible by `7` for `n=k+1`. But we initially showed that `U_n` was divisible by `7` for `n=0` and so it must also be true for `n=1, n=2, n=3, ...` and so on.

Hence, by the process of mathematical induction the given result is true for `n in NN` QED

Answer 3

See below a non inductive proof.

Explanation:

A non inductive proof

`(a^(2n+1)+b^(2n+1))/(a+b) = a^(2n)-b a^(2n-1)+b^2a^(2n-2)+cdots+b^(2n)`

or

`(a^(2n+1)+b^(2n+1))=(a+b)(a^(2n)-b a^(2n-1)+b^2a^(2n-2)+cdots+b^(2n))`

Making `a=5` and `b=2` we conclude that

`(5^(2n+1)+2^(2n+1))equiv 0 mod 7`