# Prove that equation n^a + n^b = n^c has no solutions in positive integers for n>2?

Relevance

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.

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.

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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.

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.

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

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.

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!

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.

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:

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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).

http://en.wikipedia.org/wiki/Fermat's_last_theorem

http://math.stanford.edu/~lekheng/flt/wiles.pdf