### 8 Answers

- knashhaLv 51 decade agoFavorite Answer
First we have c>a and c>b which forces a=b since otherwise

for b>a we would then have n^(b-a) divides 1. So with

a=b we divide the equation by n^a and obtain,

2 = n^(c-a) forcing (c-a)=1

and n=2 contradicting hypothesis that n>2.

Therefore there are no solutions in naturals.

- Log in to reply to the answers

- UnknownDLv 61 decade ago
Divide by n^a on both sides

1 + n^(b - a) = n^(c - a)

n^(b - a) + 1 = n^(c - a)

We can factor this.

(n + 1)(n^(b - a - 1) - n^(b - a - 2) ... ) = n^(c - a)

Let n^(b - a - 1) - n^(b - a - 2) ... = S

S(n + 1) = n^(c - a)

S = n^(c - a) / (n + 1)

S should be an integer if a, b, c, n are integers.

Since n^(c - a) only has n as a factor, n + 1 cannot be a factor.

If n + 1 is not a factor, S is not an integer, therefore a, b, c, n cannot be integers.

-----

EDIT:

Due to Lobosito's edit, I'll edit my answer.

n^(b -a) = n^(c - a) - 1

We can factor the RHS and then from division by n - 1, we show that a, b, c, n are not integers.

- Log in to reply to the answers

- Scythian1950Lv 71 decade ago
c has to be greater than either a or b. Let b > a. Then we have:

1 = n^(c-a) - n^(b-a)

c-a cannot equal b-a, otherwise the right side would = 0. Therefore, the right side has n as a factor, which cannot divide into 1 on the left side.

Please hurry up and give me my 10 points before other people give more brilliant answers.

- Log in to reply to the answers

- Theta40Lv 71 decade ago
you take the minimum of a,b,c , say a, and divide by n^a

So you get 1+n^(b-a) = n^(c-a)

suppose a is different than b, c different than a,

it implies n divides 1 , contradiction

So either a=b or c=a

If a=b, then 2=n^(c-a) , so n=2, c=a+1

contradiction

if c=a then n^(b-a)=0, contradiction

the other cases( b or c minimum) are similar.

edit: for D.L.Dennis: you can't factor

n^(b-a)+1 unless b-a is odd integer.

- Log in to reply to the answers

- What do you think of the answers? You can sign in to give your opinion on the answer.
- 1 decade ago
Just a note. This is NOT Fermat's last theorem. Look closely. Alexander isn't going to ask us to type out a 100+ page proof using only text!

- Log in to reply to the answers

- 1 decade ago
Solving it a different way:

c-1 >= a, ergo n^(c-1) >= n^a

c-1 >= b, ergo n^(c-1) >= n^a

For all n>2, ergo

n^c >2*n^(c-1) >= n^a + n^b

Cute.

- Log in to reply to the answers

- 1 decade ago
ah, good call 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10

I am sorry about that blunder, I should have read it more carefully.

Original note:

-----------------------------------------------------------------------------------

hahaha... good one.

For anyone wondering, this is the famous Fermat's Last Theorem. The theorem was one of the most famous unsolved mathematical mysteries of all time until Andrew Wiles, a professor at Princeton University, proved it in the 1990's.

Although there are many claimed proofs of this theorem, Wiles' proof is the only one to receive general recognition after many years of scrutiny (he actually had to revise the original proof he had to get it completely right).

For further information about this, follow the following links.

- Log in to reply to the answers

- dodgetruckguy75Lv 71 decade ago
Counter example:

Let n=2, a=b=1 and c=2. So 2^1 + 2^1 = 4 = 2^2

- Log in to reply to the answers