Question

find the integer solutions of \(px^2 + 2 = y^2\) where \(p\) is an odd prime

Answer 1

i found one by triall and error
\(p=7\\ x=1\\ y=3\)

Answer 2

I have a feeling this is another proof ~.~ Haha.

Answer 3

Binomial? Fermats? The ones you've been working with?

Answer 4

i have no clue yet lol still trying...

Answer 5

wolfram rejects this straight away
http://www.wolframalpha.com/input/?i=solve+px%5E2%2B2+%3D+y%5E2+over+integers

Answer 6

x=1:
p = y^2 -2

y=5, p =23
y= 7, p = 47
y=9, p =79
...
y must be odd in that case

Answer 7

Ahh thats interesting so it seems x = 1 produces infinitely many solutions as there are infinitely many primes of form y^2-2

thats neat :)

Answer 8

any easy way to conclude there are no solutions when \(x \ne 1\) ?

Answer 9

\(y^2-2\)
should not give out a prime when x not =1

Answer 10

when divided by a perfect square***

how do we convert that into a mod equation? :O

Answer 11

and why are we assuming there are no solutions when x not =1
there could be... ?

Answer 12

there is no easy way to work it wid mods i guess with squares...

Answer 13

yeah earlier i thought there can be only few solutions but after looking at ur reply it is clear that there are infinitely many solutions when x = 1

for x > 1 there could be solutions yeah.. idk i got this equation while attempting below problem which has only two solutions
http://openstudy.com/study#/updates/54b12625e4b0dbc68afe844d

looks i messed up somewhere hmm

Answer 14

im trying to find integer solutions of below equation :
\[\frac{2^{p-1}-1}{p} = n^2\]

which can be simplified like below wid some algebra
\[\frac{2^{p-1}-1}{p} = \frac{ (2^{\frac{ p-1 }{ 2 }}-1)(2^{\frac{ p-1 }{ 2 }}+1) }{ p } = n^2\]

\[ (2^{\frac{ p-1 }{ 2 }}-1)(2^{\frac{ p-1 }{ 2 }}+1) = pn^2\]

Answer 15

from there i got the equation we have been working with in this thread

Answer 16

\[ (2^{\frac{ p-1 }{ 2 }}-1)(2^{\frac{ p-1 }{ 2 }}+1) = pn^2\]

since gcd of factors on left hand side is 1 we get below equations :

\(2^{\frac{ p-1 }{ 2 }}-1 = px^2\)
and
\(2^{\frac{ p-1 }{ 2 }}+1 = y^2\)

subtracting them i get \(px^2+2 = y^2\)

Answer 17

i expected to get only two solutions to this equation because the original equation \(\frac{2^{p-1}-1}{p} = n^2 \) has only two solutions.. idk whats wrong with my work above :/

Answer 18

ur right , two solution according to Legendre theorem will work on this at home

Answer 19

mostly Diophantine for quadratic could work

Answer 20

for x=1 or -1 primes of the form y^2-2 works

Answer 21

when x is not 1 y^2-2/x^2 is not integer this is what we need to show :O

Answer 22

\(y=\pm 3,\pm 5,\pm 7,\pm 13,\pm 15,\pm 19.....\)

Answer 23

interesting...

Answer 24

>.<

Answer 25

hmm i see my silly mistake

Answer 26

hmmm well i liked that only x=1

Answer 27

great work in here

Answer 28

This reminds me of a problem I think either @No.name or @quickstudent posted a while back:

p, q, and r are all primes, what are all the solutions to this equation: p^2-q^2 = r

The method we figured out was that we could factor it to (p+q)(p-q) and since r is prime, that means one of these two has to be equal to 1, so that means p-q=1 since there's no way the other term can add two primes to get 1 haha. So then from this equation we notice p=1+q, but this implies they are consecutive primes! This only happens for q=2 and p=3. So we can check and see that 9-4 = 5 which is prime, so there is really only a single answer.

I don't know, I'm kind of sick of Fermat and Goldbach right now lol.

Answer 29

I saw you and @No.name solving it :)
I made a really silly mistake and ended up with this equation, i figured out the mistake very late... but it was fun trying to find the solutions of this equation xD

Answer 30

this is the error :
http://gyazo.com/790f52d083ea95a54f5dade23a0f2d45

:'(

Answer 31

I might be on to something! Since we have that 2 there we know x and y are either both even or both odd. So we can represent them with this formula where a is either 0 or 1. \[\Large p(2n+a)^2+2 = (2m+a)^2\] AHA! Now look and see that if a=0 this implies both numbers are even. However this leads to a contradiction: \[\Large p(2n)^2 + 2 = (2m)^2 \\ ... \\ \Large 2pn^2 + 1 = 2m^2\] Obviously since both x and y have to be even or both odd, having them both even leads to only false statements that an odd number is equal to an even one! Now we know a=1 and we can plug that in...

Answer 32

I just did a deep burial and found some answers. =)

Answer 33

Let's start digging, shall we? We know that x and y are odd and we also know that p is odd, so p is of this form, p=2k+1, so I'll just selectively plug it in where p is added rather than multiplied since it won't matter since the bulk of the following arguments are around the idea that even*even and even*odd are both even, so we have to assume the added terms are even as well:

\[p4(n)(n+1) + (p+1) = 4(m)(m+1) \\ p = 2k+1 \\ p4n(n+1) + 2k+2 = 4m(m+1) \\ p2n(n+1)+k+1=2m(m+1) \\ k=2h+1 \\ pn(n+1) + h+1 = m(m+1) \\ h = 2r+1\]

This is where the reasoning stops, we know that h has to be odd because on either side we have two consecutive numbers multiplied together which means we have an even times an odd, so the number added (h+1) must have h odd so that it can give an even result. Substituting back up the hole we have: \[\LARGE p = 8r+7\]

Answer 34

@ganeshie8 @Marki I think you'll like this. =D

Answer 35

On cursory reading, that looks like an amazing finding XD
the primes that satisfy this equation must be of form \(8k-1\) is it ?

Im going trying to read the replies thoroughly now..

Answer 36

The primes must be of the form p=8r+7 where r is any integer, so we get p=7 when r=0 like you found. =D

Answer 37

r is any positive integer*

Answer 38

that means the primes cannot be of form 8k+1/8k+3/8k+5
thats really a neat result, im still going through the replies xD

Answer 39

That doesn't necessarily mean that they all work, I just showed they must be of this form. Upon doing the wolfram alpha test we find: http://www.wolframalpha.com/input/?i=15x%5E2%2B2%3Dy%5E2+integer+solutions

Answer 40

your first reply nicely concludes that x and y must be odd xD
going thru second reply..

Answer 41

Im very tired but i liked it XD

Answer 42

really cute @Kainui :) ill save the number theory tools to prove \(p \equiv -1 \pmod 8\) for now xD

Answer 43

Here is another way to get to that result w/o using so much of number theory :

Any odd integer is of form \(4k\pm 1\)
so the equation \(px^2 + 2 = y^2\) turns out to
\[p(4m\pm1)^2 + 2 = (4n\pm 1)^2\]
\[p(8x + 1) + 2 = (8y+ 1)^2\]

dividing by \(8\) we get
\[p(1) + 2 \equiv 1 \pmod {8}\]
\[p \equiv -1\equiv 7\pmod {8}\]

this is exactly same as your method haha!

Answer 44

if we want to use number theory sophistication, there is this pell's equation which im scared of messing wid..

Answer 45

http://mathworld.wolfram.com/PellEquation.html

Answer 46

Haha interesting! I am going to have to take a break, I am tired too. But I will leave one last cute bit of nonsense, we could also rewrite the consecutive terms as if it was a sum of numbers like this to divide out one more set of 2's. \[\Large n(n+1) = 2 S_n = 2 \sum_{k=1}^n k\]

Answer 47

you're messing wid triangular numbers haha

Answer 48

I prefer to call triangular numbers "baby gauss numbers" since they always remind me of that story how he added them all up lol.

Answer 49

lol anythign to do with sum of first n natural numbers remind me the same xD
nth triangular can be given by below short form

\[S_n = \binom{n+1}{2} \]

Answer 50

using "t" for triangular is more appropriate i feel

\[t_n = \binom{n+1}{2} \]

Answer 51

\[t_{n-1} + t_n = n^2\]

greeks were really excited about properties of these numbers

Answer 52

It's kind of obvious geometrically, so I can see how they would have discovered this since algebra kinda makes it look cooler than it really is haha.