Question

A little, cute and nice question. Give it a try.

Number theory problem.

Answer 1

Prove that \(\large{2^p + 3^p}\) is not a perfect power if p is a prime number.

Answer 2

Use Mathematical Induction.. ?

Answer 3

@ganeshie8 , @mathslover , @Miracrown

Answer 4

Nope you can't use mathematical induction here.
Because you don't have a series or sequence of numbers which are prime numbers.

Answer 5

^^ The very basic criterion required to apply PMI.

Answer 6

u mean u cant write
\(2^p+3^p=(2+3)^n\) :D

Answer 7

hmm, write 3 as 2 + 1...

Answer 8

"A little, cute and nice question"

>cute

Lol

Answer 9

Yeah @ikram002p is right

Answer 10

binomial thm ^

Answer 11

You get :

\(2^ p + (2+1)^p \)

Answer 12

@marissalovescats : Of course its cute. See, how innocent is this question. It only has a 2, 3 and n. No, x or y etc.

Answer 13

Use binomial theorem now..

yeah ikram, u r right.

Answer 14

:P

Answer 15

Well guys go on give it a try with binomial theorem. Its just that I am not going to come to your rescue. :P

Answer 16

Isn't that a theorem , it says that no power of a prime is a perfect number

Answer 17

@marissalovescats : Have you ever seen a more cuter problem ? :P

Answer 18

Well, then i know to prove that

Answer 19

Nope:

2^4 + 3^2 = 25 = 5^2

Theorem fails

Answer 20

If I ever see a math problem involving cats, I'll let you know :P

Answer 21

:P

Answer 22

meow @marissalovescats :P

Answer 23

i saw a panda in p
totaly cute <3

Answer 24

I feel ya. It's always good to feel meow at 3:30am

I wear a panda necklace daily. His name's Marty. He understands me.

Answer 25

Hey that was my next question @aajugdar . You spoiled it man. XD

Answer 26

I meant no perfect number is a power of a prime.

Answer 27

hehehe

Answer 28

Okay jokes apart friends. try to prove the theorem true for one exceptional prime number first.

Answer 29

2^2 + 3^2 = 4 + 9 = 13..........! (Exceptional prime number)

Answer 30

I'll never know :(

"Three guests check into a hotel room. The clerk says the bill is $30, so each guest pays $10. Later the clerk realizes the bill should only be $25. To rectify this, he gives the bellhop $5 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn't know the total of the revised bill, the bellhop decides to just give each guest $1 and keep $2 as a tip for himself. Each guest got $1 back: so now each guest only paid $9; bringing the total paid to $27. The bellhop has $2. And $27 + $2 = $29 so, if the guests originally handed over $30.

What happened to the remaining hour?"

Answer 31

okk so forget bi
2^p+3^p=(2+3)^n
=5^n
>.>

Answer 32

so if i proved that 2^p+3^p
can devide any number not only 5 then done

Answer 33

@marissalovescats the way of thinking is wrong in this one
they took different perspective to trick you
you will get it once you actualy try it by urself

Answer 34

I've spend much time in my life trying to figure it out

Answer 35

hehehe
27 is the charge they are paying
extra 2 Rs where 25 is actualy charge :P
so 27-25 = 2 $ which the guy took :D

Answer 36

perspective :P

Answer 37

you are considering 2 Rs for the 3 Rs he gave them back xD

Answer 38

@ikram002p try FLT

Answer 39

im wondering where is it ,
its abt

n^p-1=1 mod p

if gcd (p,n) =1 right?

Answer 40

its going to be an one line proof :P

Answer 41

haha ohk
i got it nw

2^p+3^p mod p = 2+3 mod p then what ?

Answer 42

@marissalovescats simple way to know would be
30 rs is what they paid
he gave them back 3 rs
30-3 =27
n he took 2 Rs for himself
27-2 = 25
but for them they paid 30 rs for which they got 3 Rs back
so it will be
27+3 and not 27+2
tricky tricky

Answer 43

\[\large 2^p \equiv 2 \mod p \implies 2^p = pu+ 2\]
\[\large 3^p \equiv 3\mod p \implies 3^p = pv +3\]

Answer 44

set it equal to a power and show that no solutions exist

Answer 45

continue plz

Answer 46

Dang that reallly messes with my mind

Answer 47

hehe

Answer 48

Cutest OS question .. .:)

Answer 49

\(2^p+3^p=5^n\)
take mod p
\((2^p+3^p) \mod p=5^n \mod p\)

\((2+3) \mod p=5^n \mod p\)
\(5 \mod p=5^n \mod p\)
so n=1 mmm thats it ?

Answer 50

Looks perfect !

Answer 51

that remind me with fermat last thm
\(a^n+b^n=c^2\) has no solution for n >2
but i guess its not proven yet ( or idk)

Answer 52

lol you're missing the fun all these days ! google up wiley

Answer 53

*wiles proof

Answer 54

wait ill check

Answer 55

watch this sequence when free :)
https://www.youtube.com/watch?v=PPOhFok6H-Q

Answer 56

lol ok ok ive seen it before haha

Answer 57

Still qtest OS qeustion 2k14

Answer 58

ok @vishweshshrimali5 what is ur proof ?

Answer 59

Everybody...!
Now help me with my question

http://openstudy.com/study#/updates/539d53a9e4b0206eed099306

Answer 60

Ok friends here is my solution :


(1) As @mathslover did, I am going to prove the given statement true for p = 2 (only even prime number and hence an exceptional one)

\(\large{2^2 + 3^2 = 13 \ne \text{perfect power}}\)

(2) Now all prime numbers more than 2 are odd.

Now by factor theorem,

\(\large{(x+a)|(x^p + a^p)}\)

Which means that (x+a) is a factor of (x^p + a^p) when p is odd.

So, \(\large{5|(2^p + 3^p)}\)

Now, for proving that this is not a perfect number, I am going to prove that it is NOT divisible by 5^2.

Now,

\(\large{(x^p + 3^p) = (x+3)(x^{p-1} - 3x^{p-2} + ... + (-3)^{p-1})}\)

Now, when x = -3 then,

\(\large{(x^{p-1} - 3x^{p-2} + ... + (-3)^{p-1})}\)

\(\large{ = (-3)^{p-1} - 3 * (-3)^{p-2} + ... + (-3)^{p-1}}\)

\(\large{ = p*3^{p-1}}\) (Since, p is odd, so, p-1 is even, p-2 is odd and so on)

\(\large{ \ne 0}\)

So, clearly, (x+3) is not a factor of :

\(\large{(x^{p-1} - 3x^{p-2} + ... + (-3)^{p-1})}\) (factor theorem)

Thus,

\(\large{(x+3)^2}\) is not a factor of \(\large{x^p + 3^p}\).

Now, by replacing x by 2, I get that,

\(\large{(2+3)^2 = 5^2}\) is not a factor of \(\large{2^p + 3^p}\).

Thus, \(\large{2^p + 3^p}\) is not a perfect power for prime p.

Answer 61

@ganeshie8 @hartnn @ikram002p @mathslover @Miracrown

Answer 62

@ikram002p what do you think ?

Answer 63

I think this can be generalised for any x and a provided that a or x are not divisible by each other.

Thus,

\(\large{x^p + a^p}\) is not a perfect power if p is a prime number and (x,a) = 1.

Answer 64

great ^_^
nice work

Answer 65

Thanks @ikram002p

Answer 66

np , ur wlc :)

Answer 67

:)

Answer 68

very clever application of factor theorem :)

Answer 69

thanks @ganeshie8