Fermat's last theorem states:

(1) $x^n + y^n = z^n$

has no solutions for x, y, z and n positive coprime integers and n > 2. An open question is whether there exists a simple proof hinted at by Fermat. If you can spot an error in my thoughts below, please point it out.

This can be rewritten as:

(2) $x^n+(x+a)^n=(x+b)^n$ with $y=x+a$ and $z=x+b,$ a and b integers, then,

(3) $x^n+(x^n+{n\choose n-1}x^{n-1}a+...+{n\choose r}x^ra^{n-r}+...+a^n)$=

$(x^n+{n\choose n-1}x^{n-1}b+...+{n\choose r}x^rb^{n-r}+...+b^n)$, or rearranging and cancelling $x^n$ on both sides

(4) $x^n-{n\choose n-1}x^{n-1}(b-a)+...-{n\choose r}x^r(b^{n-r}-a^{n-r})+...-b^n+a^n=0$

So if we think about what this would mean for the geometrically tractable case of n=3 for a cube, so substituting n=3 in (4) gives

cubes of side x, x+a and x+b

(5) $x^3-3x^2(b-a)-3x(b^2-a^2))-b^3+a^3=0$

The cube of side x = the sum of the different cuboids between the 2 largest cubes. These differences are composed of descending terms in $x^r$. By taking these terms all on to the LHS, we are saying that the RHS = $0$.

There is another way to create a cube of volume $0$, starting from a cube of side x. If we start with a cube of side x and imagine removing a length of d until x=d, then that will create a volume of $0$. We can write this as:

(6) $(x-d)^3 = 0$, expanding gives

(7) $x^3-3dx^2+3d^2x-d^3=0$

we can then compare coefficients in ${n\choose r}x^r$ between (5) and (7) as these are both expressions describing the same geometric operation of taking $x^3$ and subtracting volumes until RHS = $0$. So, for a given value of d, we have 2 variables a and b constrained by the 3 independent equations:

(8) $d=b-a$

(9) $d^2=a^2-b^2$

(10) $d^3=b^3-a^3$

Therefore this set of equations is over-constrained and has no solutions for a and b and therefore for x, y, z for n=3. With 2 variables it is possible to define 2 of the 3 sides, but then the 3rd is not definable in the integers. Therefore (1) is impossible for n=3.

We can extend this approach for the general case of n. Equations (8), (9) and (10) will always be produced for n>2 as will

$(-1)^rd^r=a^r-b^r$ (general equation)

$(-1)^nd^n=a^n-b^n$ (final equation)

Generally there will be n equations formed when comparing coefficients but always only 2 variables. 2 of the n-dimensional sides are definable, but then the remaining dimensions have no solutions in integers.

The simultaneous equations formed in comparing coefficients between (5) and (7) will always be over-constrained for n>2. Therefore (1) is impossible.

  • 2
    $\begingroup$ It's possible for over-constrained systems to have solutions. $\endgroup$
    – B. Goddard
    Feb 9, 2019 at 23:07
  • $\begingroup$ When equating the coefficients of polynomials you are trying to find if $x^n+(x+a)^n=(x+b)^n$ has a solution in the polynomial ring $\mathbb{Z}[x]$ not in the integers ? $\endgroup$
    – reuns
    Feb 9, 2019 at 23:19
  • 2
    $\begingroup$ What is your question? If your question is, "is this a proof?", then the answer is, no, of course not. If your question is "where is/are the mistake(s) in this proof attempt?", then the answer is, it's your responsibilitiy to find the mistakes in your work. $\endgroup$ Feb 9, 2019 at 23:33
  • 1
    $\begingroup$ @GerryMyerson Not that I am a friend of fake-proofs of famous conjectures. Neither do I want to support the search for errors in such attempts. Neverhteless, I do not understand your argument. Basically, the purpose of this site is to help people to improve their mathematical skills. And showing the mistakes in a proof is such a help. Your argument might be valid in the particular case, but in general, it is not. Maybe, the proof-verification-tag would be useful. $\endgroup$
    – Peter
    Feb 10, 2019 at 10:59
  • 1
    $\begingroup$ @GerryMyerson How do you know that its not the attempt of a student, just for fun/to learn something? But even if it's not, your argument doesn't make sense to me at all. $\endgroup$
    – SampleTime
    Feb 18, 2019 at 19:44

1 Answer 1


The problem is that if two polynomials of degree 3 share a common root, their coefficients do not necessarily have to be equal (as you state in (8)-(10) comparing (5) and (7)). Look at the polynomial $(x-1)(x-2)(x-3) = x^3-6x^2+11x-6$. Its coefficients are neither the same as those of $(x-1)^3$, $(x-2)^3$ nor $(x-3)^3$.


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