Question

show that \(a^n+b^n=c^n\) has no solutions in prime numbers for all integers \(n \gt 2\)



(
a, b, c can be any prime numbers
n is an integer > 2
)

Answer 1

is this fermat's last thoerm

Answer 2

related to that, but we can prove this with some clever elementary algebra.. give it a try :)

Answer 3

ok, i was thinking u mean a,b,c primes ?

Answer 4

if yes then i'll assume "c" is odd prime ,
so "a"need to be odd, and "b" even (or inverse)
which implies at least of on them even (composite)

Answer 5

just to get ourselves acquainted with the equation, maybe first try proving below equation has no solutions in prime numbers :
\[a^3 + b^3 = c^3\]

Answer 6

2 is the only even prime

Answer 7

oh i thought ur condition a,b,c>2

Answer 8

a, b, c can be any prime numbers
n is an integer > 2

Answer 9

so , when c is even prime
then
2^n=a^n+b^n
so we have 2 cases a,b both even or both odd
but note 2 is the least prime so ay case would lead to
a^n+b^n>2^n

Answer 10

i'll rewrite in neat way

Answer 11

you're on right track!

Answer 12

case 1 :-
\(c\) is even prime , \(c=2\)
\(a^n+b^2=2^n \rightarrow \) but \(2\) is the least prime \(2^n < a^n+b^n\)

case 2:-
\(c \) is odd prime , \(c>2\)
W.L.O.G \(a \) or \(b\) need to be even and the other is odd
let \(a \) be even and \(b\) be odd
\(2^n=c^n-b^n\)
we wanna show
\(c^n-b^n > 2^n \)

Answer 13

that works!

Answer 14

An Honorary Professor of Mathematics is asking US a question?

Answer 15

Or are you teaching

Answer 16

now i'll show induction xD
\(c\) and \(b\) are both odd so \(c-b \ge 2\)
so for \(n=3\)
\(c^3-b^3>2^3\) from ur hint
now lets assume its true for k
\(c^k-b^2>2^k\)
show for k+1
\( c^{k+1}-b^{k+1} > 2^{k+1}\text{note that c is the largest btw them} \)
\(\begin{align*}
c^{k+1}-b^{k+1}&=c(c^k)-b(b^k) \\
&> c(c^k)-c(b^k) \\
&>c(c^k-b^k) \text{ use induction step} \\
&>c.2^k \text{note c>2} \\
&>2^{k+1}
\end{align*}\)

Answer 17

Tadaaa

Answer 18

so this work when u prove
\(a^3+b^3=c^3\) has no solution
i assumed it worked , but if u wanna we can go through it maybe

Answer 19

Excellent! looks neat.

Answer 20

\(\text{for } a^3+b^3=c^3 \text{ we only need to show} \\c^3-b^3>2^3\\\begin{align*}
c^3-b^3 &= (c-b)(b^2+bc+c^2)\\
&>2.2^2 \\
&>2^3
\end{align*} \)

Answer 21

if we assume \(c\ne b \gt 2\) we have :
\[\begin{align}c^n - b^n &= (c-b)(c^{n-1} + c^{n-2}b + \cdots + b^{n-1}) \\~\\
&\gt 2(c^{n-1} + c^{n-2}b + \cdots + b^{n-1}) \\~\\
&\gt 2(2^{n-1}) \\~\\
&= 2^n
\end{align} \]

Answer 22

that proves that there are no solutions when atleast one of a,b,c is 2

Answer 23

hehe i was looking for this expansion >.< i

Answer 24

yes yes ! short and neat

Answer 25

lol @CheesecakeKitten i just thought this is an interesting problem because it refers to Fermat's last theorem

look it up in google, you will find fascinating stories im sure :)

Answer 26

Fermat was for integers right ?

Answer 27

I'd rather not read about math for fun, thank you.

Answer 28

yes http://mathworld.wolfram.com/FermatsLastTheorem.html

Answer 29

WAY too complicated

Answer 30

srry!

Answer 31

and i'm kinda supposed to be doing my school work

Answer 32

@CheesecakeKitten you dont have , just dont waste ur time darling ;)

Answer 33

have to*

Answer 34

i can't help it, I'm a procrastinator. :P

Answer 35

lol

Answer 36

haha

Answer 37

-.-

Answer 38

omfg i'm being spammed with messages!!!

Answer 39

i spammed u :O

Answer 40

hehe will go now cya ;)

Answer 41

lol funny pic

Answer 42

-.- yeah

Answer 43

this is my fav

Answer 44

XD

Answer 45

http://www.puravidamultimedia.com/wp-content/uploads/2013/09/o_102096.jpg

Answer 46

thats hilarious haha (will type the rest later ;) )

Answer 47

XD

Answer 48

my profile pic. c:

Answer 49

just changed it to this an hour or so ago.

Answer 50

https://www.youtube.com/watch?v=LOMbySJTKpg

Answer 51

cool

Answer 52

can you help me @ganeshie8