Prove that equation n^a + n^b = n^c has no solutions in positive integers for n>2?
- knashhaLv 51 decade agoFavourite 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.
- 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.
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.
- 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.
- 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
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.
- 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!
- 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
- 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.
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.
- dodgetruckguy75Lv 71 decade ago
Let n=2, a=b=1 and c=2. So 2^1 + 2^1 = 4 = 2^2