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Fermat Last Theorem look-alike

  1. Nov 6, 2003 #1
    I encountered this problem when I was reading Dr Riemann's Zero, the problem is:

    Find the integers a, b and c such that a^3 + b^3 = 22c^3.

    First I did some research from the public library and obtained some lecture notes on Fermat's Last Theorem, but I made little progress, partly because I don't understand most part of the book.

    The only thing I know about this equation is that they are all relative primes, and a^3 and b^3 must be odd numbers.

    Can somone please show me how this is done?
     
  2. jcsd
  3. Nov 7, 2003 #2

    Hurkyl

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    A lot of times, when I've seen diophantine equations solved directly, one plugs what they know back into the equation and tries to elicit new information.

    In this case, you know a and b must be odd, so write a=2d+1 and b=2e+1, expand, simplify, and see if you can discover something else.



    (p.s. are you searching for solutions or trying to prove impossibility?)
     
  4. Nov 25, 2003 #3
    sorry, I tried to solve it. I'll try again someday.
     
    Last edited by a moderator: Nov 25, 2003
  5. Nov 25, 2003 #4
    Re: Re: Fermat Last Theorem look-alike

    it should be: 7a^3+7b^3+21ab^2+21ba^2=ab^2+ba^2
     
  6. Nov 25, 2003 #5
    Re: Re: Re: Fermat Last Theorem look-alike

    I corrected that in a few minutes, you're a fast commenter.
     
  7. Jul 10, 2004 #6
    Need ideas! Any ideas?

    Cubic Problem X^3 + Y^3 = 22Z^3 a problem.

    Considering the method of Euler-http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html#40-
    we are able to achieve the form:

    (X+Y) (X^2-XY + Y^2) = (2m) (m^2+3n^2) by setting X = m +n, Y =m-n.

    This form is closed under multiplication, that is:

    (s^2+3t^2)(u^2+3v^2) = (su-3tv)^2 + 3(ut + sv)^2,

    We now cube the form a^2+3b^2. This results in:

    [(a^2-3b^2)a-3b(2ab)]^2 + 3[(a^2-3b^2)b + a(2ab)]^3

    Setting this equal to m^2 + 3n^2 and simplifying we have a cube where:

    m = a(a^2-9b^2). n= 3b(a^2-b^2)

    Example: Let us suppose a=1, b=2, then m=-35, n=-18. Thus:

    m^2 +3n^2 =35^2 +3x18^2 = 2197= 13^3.

    While a=2, b=3 leads to

    154^2 + 3x45^2 = 29791 = 31^3.

    Now we want to take in consideration 2m, putting it all together for a=2, b=1, we get m=-10, n=9, 2m=-20, X=-1, Y=-19.

    X^3 +Y^3 = 2m(Z^3) = 1^3 + 19^3 = 20(7^3).

    But we really want to find 2m = 2x11xv^3. But this is a real problem. Some results are found: 51^3 + 111^3 = 6x63^3, and 152^3 + 8^3 =20x56^3, and 67881^3 + 147741^3 = 6x(7623x11)^3.

    By using a=8, b=1, we can get m==0 Mod 11, but this also satisfies a == 3 Mod 5. This approach gives us a “near miss”:

    629^3+251^3 = 110(134)^3.

    I am out of ideas.
     
  8. Jul 10, 2004 #7

    Gokul43201

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    Haven't really read all the posts here but here's something that may or may not be useful.

    If solutions exist, then a,b would have to be of the form 11k + n, 11m - n with n in [1,10].

    This results from solving [tex]~~a^3 + b^3 \equiv 0 ~~(mod~11)[/tex]
     
    Last edited: Jul 10, 2004
  9. Jul 10, 2004 #8
    I've seen this discussed on sci.math, I don't think anyone there had a solution though...
     
    Last edited: Jul 10, 2004
  10. May 13, 2009 #9
    I ran into this problem yesterday in the introduction to The Riemann Hypothesis: The Greatest Unsolved Problem in Mathematics by Karl Sabbagh. I think I just solved it, but if I've made a mistake somewhere, I'd appreciate someone pointing it out.

    The version of the problem I have is: Find positive integers a, b, and c such that a^3 + b^3 = 22c^3

    I took the approach of Gokul43201, except that I centered both (11k + n) and (11m - n) around the same multiple of 11. Thus, in my notation, a = 11x - k, b = 11x + k, where k can be greater than 11. Also, b - a = 2k, which is needed to assure that a^3 + b^3 is even.

    So, we have:
    (11x - k)^3 + (11x + k)^3 = 22c^3

    Expanding and collecting terms, we have:

    121x^3 + 3xk^2 = c^3

    From this, we can see that x must be a factor of c. That is, there must exist positive integer y such that xy = c. Thus, we have:

    121x^3 + 3xk^2 = (xy)^3
    3xk^2 = x^3 * y^3 - 121x^3
    3k^2 = x^2 * y^3 - 121x^2
    3k^2 = x^2(y^3 - 121)

    From this, we can see that x^2 must be a factor of 3k^2, and x must be a factor of k. There must exist positive integer m such that mx = k. Therefore, we can reduce to:

    3m^2 = y^3 - 121

    We've eliminated a variable (is this problematic? I was hoping not).

    Now, the right side must be divisible by 3. Since 121 (mod 3) = 1, then y^3 (mod 3) = 1, and there must exist positive integer z such that (3z + 1) = y. Thus,

    3m^2 = (3z + 1)^3 - 121
    3m^2 = 27z^3 + 27z^2 + 9z - 120
    m^2 = 9z^3 + 9z^2 + 3z - 40
    m^2 + 40 = 3z(3z^2 + 3z + 1)

    From this, we see that m^2 + 40 must be divisible by 3. Since 40 (mod 3) = 1, that means m^2 (mod 3) = 2. But this is impossible. Consider, if x (mod 3) = 1, then x^2 (mod 3) = 1. If x (mod 3) = 2, then x^2 (mod 3) = 1. Therefore, there exist no positive integers a, b, and c such that a^3 + b^3 = 22c^3

    This is the first time I've attempted a proof of this sort, so I've probably made a mistake somewhere, and am looking forward to being corrected.

    Thanks,
    Erich Enke
     
  11. May 13, 2009 #10

    Hurkyl

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    In case you missed it, you needed to know that b-a is even in order to center them both around the same multiple of 11. e.g. if a=1 and b=10, they are of the form a=11k-n and b=11m-n, but cannot be written in the form you used.
     
  12. May 13, 2009 #11
    Correct. I think that was in my reasoning process, but I reported it a bit awkwardly. To put it in other words, since a^3 + b^3 must be even, the difference between a and b must be even. a and b therefore cannot be (11x + 1) and (11x + 10) respectively, since their difference will always be odd.
     
  13. May 13, 2009 #12
    Hm.. but I do agree with you -- that's a bit of a weak point in the proof. Am I handling all the cases when I center around the same multiple of 11? Or are there some legitimate cases I'm not considering, such that I'm only proving that there is merely no solution within a subset of possible solutions? I'll give some thought to that.

    Thanks,
    Erich
     
  14. May 13, 2009 #13
    Ok, how about this:

    Assume a = (11d + n) and b = (11e + 11-n). Knowing the difference of a and b must be even, then:

    (11d + n) - (11e + 11 - n) = 2k
    11(d - e - 1) = 2(k - n)
    d - e must always be odd.

    Instead of using e, if we choose f such that we can use -n instead of 11-n, then we have

    (11d + n) - (11f - n) = 2k
    11(d - f) = 2(k-n)
    d-f is always even.

    If d-f is even, there always exists an integer midpoint m between d and f such that m=(d+f)/2 and d-m = (d-f)/2 and m-f = (d-f)/2

    Therefore, we can always write (11d + n) - (11f - n) = 2k in terms of m and a radius r=(d-f)/2:
    11(m+r) + n - 11(m-r) - n = 2k

    since m+r = (d+f)/2 + (d-f)/2 = d and m-r = (d+f)/2 - (d-f)/2 = f

    Therefore, every considerable pair of (mod 11) numbers can be represented as equidistant from a central multiple of 11.
     
  15. May 13, 2009 #14

    Hurkyl

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    Shorter derivation: the desired multiple of 11 is the average of a and b, which is an integer iff their sum is even.
     
  16. Dec 8, 2009 #15
    17,299^3 + 25,469^3 = 22(9,954^3), both 17,299 and 25,469 are prime.

    I, too, stumbled on this problem a couple years ago. I tried all the methods listed in this thread without finding a closed solution. There was no magic finding this answer. I wrote a program for my HP48 calculator to search for it. Of course, what you'd like is a solution similar to the pythagorean progblem:

    Find integers that solve a^2 + b^2 = c^2.
    Answer: (a - b)^2 + 4ab = (a + b)^2 => (a - b)^2 + (2(ab)^.5)^2 = (a + b)^2
    Now, you can choose any a and b you like so long as the square root of ab is an integer. But, this is easy. Choose any set of prime factors you like. Raise them all to even powers. Partition them between a and b in any manner whatsoever. Poof! You're done.
     
  17. Dec 8, 2009 #16
    I'd like to correct a slight booboo in my previous. There's no change to the a^3 + b^3 = 22c^3 solution. But below in the Pythagorean problem, I used 'a' and 'b' for two different purposes. See below.

    17,299^3 + 25,469^3 = 22(9,954^3), both 17,299 and 25,469 are prime.

    I, too, stumbled on this problem a couple years ago. I tried all the methods listed in this thread without finding a closed solution. There was no magic finding this answer. I wrote a program for my HP48 calculator to search for it. (The program ran for 10 days or so and exhausted at least 4 sets of batteries.)

    Of course, what we'd like is a solution similar to the pythagorean problem:

    Find integers that satisfy x^2 + y^2 = z^2. (This is the correction. I now use variables x, y, and z. Previously I used a, b, and c which were not intended to be the same as 'a' and 'b' in the answer.)

    Answer:
    (a - b)^2 + 4ab = (a + b)^2
    (a - b)^2 + (2(ab)^.5)^2 = (a + b)^2

    Now, you can choose any a and b you like so long as the square root of ab is an integer. But, this is easy. Choose any set of prime factors you like. Raise them all to even powers. Partition them between a and b in any manner whatsoever. Poof! You're done.
    Example
    set of primes = [3(5)(7)(11)]^2
    A = (5^2)(7)(11) = 1925
    B = (3^2)(7)(11) = 693
    A+B = 34(7)(11) = 2618
    A-B = 16(7)(11) = 1232
    2(AB)^.5 = (2)(3)(5)(7)(11) = 2310
    (a-b)^2 + (2(ab)^.5)^2 = (a+b)^2 <=> 1232^2 + 2310^2 = 2618^2

    Also, Mr. Erich Enke's (post #9) proof goes awry when he sets xy = c. He says that x must be a factor of c. In fact, all we can say is that x must be a factor of c^3. Thus, instead of setting xy = c, he should've set xy = c^3. Again, the prime factors of x need not be contained within c, only within c^3.
     
  18. Jun 4, 2010 #17
    Hello everybody,
    This is my first post in this forum. I would like to write a demonstration of the Fermat Theorem I developed in late 1993, put in a floppy and then forgot for almost 17 years. Now I found it again and want to spread all over the web to see if it is valid. If this is the case I invite you to send me a message to g_pet@hotmail.com.
    Well let's go (forgive my italianish english).

    Hp: x,y belong to rationals (x,y <> 1 and 0)

    n belongs to naturals; n>=3



    Th: x^n + y^n <> 1



    Demonstration by absurd. Let's suppose that it exists at least one couple
    x and y so that x^n + y^n = 1.



    x^n + y^n = 1 <=> x^n = 1 - y^n <=> x = (1- y^n)^(1/n)



    since y belongs to rationals so it can be written as

    y = p/q with p and q belonging to integers.
    let's suppose that they are also primes between them otherwise we simplify the common factors
    . From this it comes that:

    x = (1 - (p^n)/(q^n))^(1/n) <=> x = (((q^n)-(p^n))^(1/n))/(abs(q))



    At this point we can note that also ((q^n)-(p^n))^(1/n) will belong…

    to rationals because it equals x*(abs(q)) which is itself a rational

    Therefore ((q^n)-(p^n))^(1/n) has to be of the form a/b with a and b
    primes between them.



    ((q^n)-(p^n))^(1/n) = a/b <=> q^n - p^n = (a^n)/(b^n)



    From which


    (q*b)^n - (p*b)^n = a^n <=> a^n + (p*b)^n = (q*b)^n



    Here is the absurd. In fact the number on the right contains in its
    decomposition in primes factors the factor q: now for the Euclide's
    theorem on the uniqueness of the decomposition in primes factors
    also the number on the left has to contain the factor q
    which is
    not possible because at most two cases should present, both absurd


    1) p contains in his decomposition the factor q: this is against the
    hypothesis that p and q are primes so this possibility has to be
    thrown away

    2) b contains in his decomposition the factor q: in this case also the factor
    a has to contain, in his decomposition the factor q and thus a and b could
    not be primes between them: also this has to be thrown away because
    of the previous hypothesis

    At the end the negation of the thesis is absurd and therefore it is demonstrated

    Ciao
    Gianluca Pettinello
     
  19. Jun 4, 2010 #18

    Hurkyl

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    The same argument would conclude that 1+2=3 is absurd....
     
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