Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - Prove this is the only solution

  1. #1
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56

    Prove this is the only solution

    a^2 + 2 = b^3
    a=5 and b=3
    Prove that a=5 and b=3 are the only solutions.

    Any ideas?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Ruichan View Post
    a^2 + 2 = b^3
    a=5 and b=3
    Prove that a=5 and b=3 are the only solutions.

    Any ideas?
    There must be some restriction on what constitutes a solution missing here,
    as otherwise a=-5, b=3 is also a solution. Perhaps you are interested in
    solutions in natural numbers?

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56
    Quote Originally Posted by CaptainBlack View Post
    There must be some restriction on what constitutes a solution missing here,
    as otherwise a=-5, b=3 is also a solution. Perhaps you are interested in
    solutions in natural numbers?

    RonL
    I have no idea, my prof gave us this game to play, this is the only detail he gave us. He said those are the only solutions to the above. a = 5 and b = 3
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Ruichan View Post
    I have no idea, my prof gave us this game to play, this is the only detail he gave us. He said those are the only solutions to the above. a = 5 and b = 3
    What were you studying in class just before he set this?
    (it could be a clue to what restrictions on solutions are intended)

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56
    Quote Originally Posted by CaptainBlack View Post
    What were you studying in class just before he set this?
    (it could be a clue to what restrictions on solutions are intended)

    RonL
    Actuarial Math but i don't think it's related.
    Other stuffs he sort of provided are:

    (a^2 -25) = b^3 - 27
    <br />
(a + 5)(a - 5) = (b-3)(b^2+3b+9)<br />
    a (subscript k) +5 = k^2 +10k

    1.(0,37) = 37
    2.(37,98) = 61
    3. (98,189) = 91
    4. (189,316) = 127

    2.-1. = 24
    3.-2. = 30
    4.-3. = 36

    He did say that he proved it long time ago, nobody else, at least those he knows could prove it.

    I actually have no idea what's he talking about though.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    In the 17th Century my Favorite mathemation Pierre de Fermat wrote that 26 is the only number contained in between a square and a cube.
    That leads to the equation,
    x^2+2=y^3.
    Fermat never wrote down the proof (like usual) and many people failed to prove it (including me ) but about 100 years later Leonard Euler proved it. Though (I think) it he spend 7 years on this problem! (Just goes to show the Fermat was the master of Diophantine Equations). I myself have no idea how to show that.

    I happen to know that x^3+1=y^2 also has no solutions accept x=2,y=3 which is (special case of Catalan Conjecure--- now proven). But was shown to be true also by Euler I think and again all texts on number theory omit the prove because it is way too long (though clever). However, I think there is a simpler proof due to Gauss based on the Quadradic fields and complex factorization of the cube which leads to an impossibility. But I do not think Gauss' method works here because the expression x^3+2 is not factorable while x^3+1=(x+1)(x^2-x+1) which is the entire argument. Finally there is certainly a way to show this using the theory of Elliptic curve because this leads to a the square root of a cubic polynomial, but sadly I am an noob so I cannot know of one.

    By the way, I really do not think your professor proved it using Elementary number theory. I can promise you that this problem is a problem only for the gods. I think he just lied.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,928
    Thanks
    333
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    In the 17th Century my Favorite mathemation Pierre de Fermat wrote that 26 is the only number contained in between a square and a cube.
    That leads to the equation,
    x^2+2=y^3.
    Fermat never wrote down the proof (like usual) and many people failed to prove it (including me ) but about 100 years later Leonard Euler proved it. Though (I think) it he spend 7 years on this problem! (Just goes to show the Fermat was the master of Diophantine Equations). I myself have no idea how to show that.

    I happen to know that x^3+1=y^2 also has no solutions accept x=2,y=3 which is (special case of Catalan Conjecure--- now proven). But was shown to be true also by Euler I think and again all texts on number theory omit the prove because it is way too long (though clever). However, I think there is a simpler proof due to Gauss based on the Quadradic fields and complex factorization of the cube which leads to an impossibility. But I do not think Gauss' method works here because the expression x^3+2 is not factorable while x^3+1=(x+1)(x^2-x+1) which is the entire argument. Finally there is certainly a way to show this using the theory of Elliptic curve because this leads to a the square root of a cubic polynomial, but sadly I am an noob so I cannot know of one.

    By the way, I really do not think your professor proved it using Elementary number theory. I can promise you that this problem is a problem only for the gods. I think he just lied.
    Crap. I was really hoping TPH would come through on this one (as he likes these) and I've now spent several hours working on it. All I can say is that the solution must be of the form:
    a = 210x + 5, b = 210y + 3
    or
    a = 210x + 5, b = 210y + 33
    (x and y integers, not necessarily positive)

    Obviously the first pair contains the (presumably only) solution and I have yet to find any solutions using the second. Unfortunately the method I am using will never generate only one solution, so it's a dead end (though an enjoyable piece of work.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by ThePerfectHacker View Post
    26 is the only number contained in between a square and a cube.
    What's that supposed to mean
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56
    Quote Originally Posted by ThePerfectHacker View Post
    By the way, I really do not think your professor proved it using Elementary number theory. I can promise you that this problem is a problem only for the gods. I think he just lied.
    I don't think he's lying though. His solution was published in reader's digest or whatever magazine. I wasn't paying attention to what he was saying coz I was practically sleeping with my eyes open by 8.30pm last night.

    Guess I just need to ask him next week on which magazine that solution was published in. However, it was published almost 2 decades ago though.

    He didn't proved the theory, just proved that a = 5 and b = 3 are the only solutions.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,928
    Thanks
    333
    Awards
    1
    Quote Originally Posted by Quick View Post
    What's that supposed to mean
    25 = 5^2 < 26 < 3^3 = 27: 26 is the only such positive integer that is in succession between a square and a cube.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by topsquark View Post
    25 = 5^2 < 26 < 3^3 = 27: 26 is the only such positive integer that is in succession between a square and a cube.

    -Dan
    Oh, now I see. I was thinking things like 3^2<10<3^3
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,928
    Thanks
    333
    Awards
    1
    I don't know, maybe someone will gain some inspiration from what I was trying and succeed where I failed. I was trying an inductive type proof that a and b could only be expressed as:
    a = \left (2 \prod_{i = 1}^n(2i+1) \right )m + 5
    b = \left (2 \prod_{i = 1}^n(2i+1) \right )q + 3
    where m and q are integers.

    If this is true and you can do an induction, then we can show that n is without bound. If this is true then the only solution is for m, q = 0. That is, a = 5, b = 3. But I couldn't manage to prove the inductive step. (I kept on finding other forms that I couldn't prove were incorrect.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Ruichan View Post
    I don't think he's lying though. His solution was published in reader's digest or whatever magazine. I wasn't paying attention to what he was saying coz I was practically sleeping with my eyes open by 8.30pm last night.
    .
    I would love to see it.
    ---
    Topsquark just drop it. You are not going to solve it. I spend hours on these diophantine equations including this one. Trust me when I say that they are not easy, they might seem easy. Especially when you introduce cubes. I have seen proofs of several diophantine equations for example Fermat's last theorem for n=3, by Euler, it is huge, just pages and pages and pages all using manipulations divisibility arguments. They also involve beyond belive algebraic manipulations and the most unusual substitution. Which is why Euler solved most of them, becuase he was one of the few that was able to do this.

    Solving diophantine equations in this type is called an "elementary" solution. My problem of the week used such an argument. There are also other non-elementary methods, one became famous in 19th century using field theory. And today there is a another one using elliptic curves. I think your professor solve it using the two latter methods, not the classical, but it would be nice if you can post it here. I would like to see how it is approach using modern math.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56
    Mathematics or Mathematical Monthly. It's from Mathematics Association of America, Journal form. That issue was published between 1980-1981.

    Prof doesn't remember how to prove it, HEHEHE, he's looking for the hard copy of his proof. He forgot where he place his hardcopy.
    That's why he's asking if we know how to prove it.

    He said he only remembered that he used the modular formula or whatever.

    I'm too lazy to go search for through the last 20 years of that issue which his proof was accredited.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Ruichan View Post
    He said he only remembered that he used the modular formula or whatever.
    Exactly, what I said, a non-elementary proof. Modular forms are 20th century mathematics. If he really knows them that is something to be proud of.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Prove x^6+7y^3=2 has no solution
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: March 5th 2011, 01:32 PM
  2. Prove solution exists
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 4th 2010, 08:29 PM
  3. Prove this solution??
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 27th 2010, 09:20 PM
  4. Prove there always be a solution
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: November 14th 2009, 04:40 PM
  5. Replies: 1
    Last Post: May 3rd 2009, 12:57 AM

Search Tags


/mathhelpforum @mathhelpforum