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Integer Solutions to a Cubic Equation

Date: 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/ 
Associated Topics:
College Number Theory
High School Number Theory

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