Integer Solutions to a Cubic EquationDate: 04/11/2005 at 12:24:16 From: Kai Subject: Roots of a cubed function If a, b, and c are all whole numbers, without a = b = c = 0, can a^3 + 2b^3 + 4c^3 - 4abc = 0? I do not know much about cubed functions so I do not know how to factor anything out. I first tried to simply the function by moving the 4abc to the opposite side. I then tried to divide both sides by abc. This just got me more confused. Date: 04/12/2005 at 14:14:37 From: Doctor Ricky Subject: Re: Roots of a cubed function Hey Kai, Thanks for writing Dr. Math! This is a pretty interesting question. In fact, it illustrates an idea that Pierre de Fermat used to prove certain cases of his Last Theorem to be true (Fermat's Last Theorem says that x^n + y^n =/= z^n for any n > 2). Our equation is: a^3 + 2b^3 + 4c^3 - 4abc = 0 where a, b, c are whole numbers and not all equal to zero. Our first step would be to add 4abc to both sides, which gives us: a^3 + 2b^3 + 4c^3 = 4abc Now, if we divide by 4, we see that: a^3 + 2b^3 + 4c^3 ----------------- = abc 4 Remember that abc is a whole number since the product of whole numbers is a whole number. Therefore, since a^3 + 2b^3 + 4c^3 is divisible by 4, we know that both a^3 and 2b^3 are divisible by 4. Note that 4c^3 is already divisible by 4. Therefore a and b must both be even. This means that a = 2m and b = 2n , for some integers m and n where m < a and n < b. If we plug this into our equation for abc, we end up with: (2m)^3 + 2[(2n)^3] + 4c^3 ------------------------- = (2m)(2n)c 4 which is the same as: 8m^3 + 16n^3 + 4c^3 ------------------- = 4mnc 4 which simplifies to: 2m^3 + 4n^3 + c^3 = 4mnc Dividing again by 4, we see: 2m^3 + 4n^3 + c^3 ----------------- = mnc 4 For this to be true, 2m^3 and c^3 must be divisible by 4 and are therefore both even. Therefore, m = 2j and c = 2k for some integers j and k, j < m and k < c. Then we plug these into our equation to get: 2[(2j)^3] + 4n^3 + (2k)^3 ------------------------- = (2j)n(2k) 4 which, upon simplification, gives us: 16j^3 + 4n^3 + 8k^3 ------------------- = 4jnk 4 and reduces to: 4j^3 + n^3 + 2k^3 = 4jnk As you might notice by now, we can repeat this process an infinite number of times without getting something different than 4x^3 + y^3 + 2z^3 = 4xyz for some integers x, y and z. Remember, however, that a = 2m, or in other words, m = a/2. Now remember that m = 2j, or j = a/4. We can show this with our other variables, also. As we keep going, we see that each successive variable substitution requires us to divide 'a' by 2 again. However, we said we can repeat this process infinitely, and there is no even integer 'a' such that if we divide it by 2 an infinite number of times it will stay an integer. Similarly we can show this for our other two variables by demonstrating the same respective relationships. Therefore, there is no solution for this equation. This method of proving is known as Fermat's method of infinite descent and is the basis idea for the common method of proof known as mathematical induction. I hope this helps! If you have any more questions, please let me know! - Doctor Ricky, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/