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Euler's (wrong) Proof of FLT for Cubes

Before you take the time to study this proof, I have an admission to make: it's wrong.  It relies on "Lemma 2", which is presented at the bottom of this page, and, alas, Lemma 2 is just plain false.  Here is the proof anyway, which is interesting for historical reasons only:

There are no non-trivial solutions to x3+y3=z3  

Euler�s Proof of the case n=3

If x, y, z is a non-trivial solution to x3+y3=z3, (one in which none of x, y, z are zero), and any two of x,y,z share a common factor, then the third variable shares that same factor, and so a smaller solution can be found in which x, y, and z are pairwise coprime.  ("Coprime" means sharing no common factor other than 1; "pairwise" means when taken in pairs.)

So let x, y and z be the solutions for x3+y3=z3 where x, y and z are pairwise coprime.  One and only one of the numbers, x, y or z must be even as they are pair wise relatively prime while the sum of two odd numbers must be even.

Since x3+y3=z3, it is true that z3+(-x)3=y3, and z3+(-y)3=x3, so a manipulation such as this can easily be used to put the two odd numbers on the left side of the equation.

WLOG (Without loss of generality), let x and y be odd, and z be even.  Then x+y and x-y are both even, say 2p and 2q respectively, and x=p+q, y=p-q.  x3+y3 can be factored this way:


p and q are of opposite parities as x=p+q and y=p-q are both odd.  They are also coprime because their common factor would also be a common factor of p+q and p-q, which contradicts that fact that x=p+q and y=p-q are coprime.

Note that neither p nor q can be zero.  If p were zero, then the solution would be trivial.  If q were zero, then x=y, and since they're coprime, they are both equal to 1, and z3=2 is false when z is an integer.

Now, WLOG, we can assume p and q are both positive.  If p were negative, then we could multiply through by -1 to "flip" the sign of every term, making p positive.  If q were negative, we could interchange x and y, making q positive.

Recapitulation of assumptions and deductions so far

x3+y3=z3, where x, y, and z are coprime integers (some may be negative)
x and y are both odd, and z is even
positive coprime integers p and q exist such that x+y=2p and x-y=2q
x3+y3 = z3 = 2p(p2+3q2)

If it turns out that 2p and p2+3q2 have no common factors, then since their product is a cube, each one must be a cube, by Lemma 1 (below).

If they do have a common factor, then what are their possible common factors?  2 can't be a common factor of 2p and p2+3q2, because p2+3q2 is odd.  Any common factor of 2p and p2+3q2 must be a common factor of p and p2+3q2, and so this must also be a factor of both p2 and 3q2.  Since p and q are coprime, the only possible common factors of p and 3q2 are 1 and 3.

Now we will consider two cases:
Case 1, in which gcd(2p, p2+3q2)=1; and
case 2, in which gcd(2p, p2+3q2)=3.

Case 1: gcd(2p, p2+3q2)=1

Since we assume that there is no common factor between 2p and p2+3q2, 2p and p2+3q2 are both cubes, by Lemma 1 (below).

There is a clever formula, similar to the Complex Product Identity, which shows that the product of two numbers, each of which has the form a2+3b2, also has the same form:


So (a2+3b2)2 is given by

(a2+3b2)(a2+3b2) = (aa-3bb)2+3(ab+ba)2  = (a2-3b2)2+3(2ab)2 

And (a2+3b2)3 is given by

(a2+3b2)(a2+3b2)2 = (a2+3b2)((a2-3b2)2+3(2ab)2)
 = (a(a2-3b2)-3b(2ab))2+3(a(2ab)+b(a2-3b2))2  
 = (a3-3ab2-6ab2)2+3(2a2b+a2b-3b3))2  
 = (a3-9ab2)2+3(3a2b-3b3))2  

Therefore, we can find cubes of the form p2+3q2 by choosing a, b at random and to set p=a3-9ab2 and q=3a2b-3b3 so that p2+3q2=(a2+3b2)3.

Unfortunately, Euler's proof has a gap: we haven't shown that this is the only way for p2+3q2 to be expressed as a cube.  That is, we haven't shown that whenever p2+3q2 is a cube that there must exist an a and b such that p and q are given by the equations p=a3-9ab2 and q=3a2b-3b3.  If this gap can be closed, then the proof is valid.  I left it, unproved, as Lemma 2, below.  Letting this slide by, for the moment, we will continue Euler's proof...

Factor the expressions p=a3-9ab2 and q=3a2b-3b3 so that

p=a(a-3b)(a+3b), and

a and b are coprime because any common factor of them will divide both p and q, which we've already established are coprime.

2p can now be expressed in terms of a and b so that 2p = 2a(a-3b)(a+3b), which is a cube.

a and b must have opposite parities.  Otherwise p and q would be both even.  As a result, a-3b, a+3b are both odd and the only possible common factor of 2a, a+3b, and a-3b would be common factors of a, a+3b, and a-3b, and therefore of a and 3b.  We can then see that the only possible common factor is 3 but 3 does not divide a because it would then divide p, hence contradicting the assumption in case 1.  Therefore 2a, a-3b, a+3b are pairwise coprime and so all three of them must be cubes.  If we let 2a=Z3, a-3b=X3, and a+3b=Y3, this gives a solution of X3+Y3=Z3.

As Z3X3Y3=2a(a-3b)(a+3b)=2p, which is a proper factor of z3, so X, Y, and Z are all smaller, in absolute value, than z.  Thus, the solution X3+Y3=Z3 uses smaller numbers than the original x3+y3=z3.

The process can continue for infinite number of times to get infinitely many smaller and smaller value of solutions for the equations.  However, for positive integers, it cannot decrease indefinitely.  An infinite descent has been set up.  Hence we can conclude that the case 1 of the case n=3 of Fermat�s Last Theorem is proved.

Case 2: gcd(2p, p2+3q2)=3

In case 2, we assume that 3 is a factor of p.  Then p=3k, for an integer, k, and 3 does not divide q.  Then 2p(p2+3q2)=32*2k(3k2+q2).  32*2k and 3k2+q2 are coprime (remember that 3 is not a factor of q) hence both of them are cubes.  Again, we have to first prove that for 3k2+q2 to be a cube, q must equal a(a-3b)(a+3b) and k=3b(a-b)(a+b) for some integers a, b. Since 32*2k is a cube, 33*2b(a-b)(a+b) is a cube and therefore 2b(a-b)(a+b) is a cube.  We can see that the factors are relatively prime.  Now take 2b=X3, a-b=Y3, a+b=Z3, and hence X3+Y3=Z3 with Z3<z3.  Once again, an infinite descent is accomplished and the theorem is proved.

Lemma 1: if gcd(a,b)=1 and ab=zk, then integers p and q exist such that a=pk and b=qk.

For each prime factor, p, of ab=zk, p is a factor of a or b, but not of both a and b.

If p|a, then pnk|a, where n is a positive integer.

Similarly, if p|b then pnk|b.

So a and b are each the product of primes raised to powers that are multiples of k, so both a and b are kth powers of an integer.

Lemma 2: If a cube, m3, can be expressed as the sum of a square and three times a square, then integers a, b exist such that p=a3-9ab2 and q=3a2b-3b3 , and m3=p2+3q2.

proof?  Alas, it isn't even true!


Commentary about the proof

From time to time I'm asked why certain things mentioned in this proof are true.  First of all, my advice is not to spend too much time on this proof, because as I point out in the first paragraph, above, the proof is flawed.

Euler's is a type of proof that's divided into "cases".  Recall that x3+y3=z3, with x and y both odd and coprime. x=p+q and y=p-q, and so (without loss of generality) p and q are both positive, coprime, and of opposite parity. Thus gcd(2p, p2+3q2) is either 1 or 3.

Sometimes people skip over case 1, because it's too long and complicated, and start by looking at case 2.  This is a mistake, because case 2 uses the trick of showing that if both 2p and p2+3q2 are multiples of 3, then you can divide them by 3 to get 2k and q2+3k2, and then all the same arguments apply, with q and k playing, in case 2, the roles that p and q played in case 1.  So you really have to go back to case 1 and read those arguments to understand them.

The explanation of how we carve p2+3q2 into a's and b's is given fully in the narrative for case 1.  And this is the crux of the flaw in Euler's proof, so I don't know how much time you really want to spend on this.  This is the kind of flaw that is really hard to spot, and once you do spot it, it's really hard to understand and fix it because the whole discussion is about something that can't happen, which is always the case in proofs by contradiction, particularly "infinite descent" proofs such as this one.  Remember, the proof is: if x3+y3=z3, then a "p" and a "q" exist with certain properties, and among these properties is the existence of "a" and "b" with these other properties. The whole thing is based on a premise that is false from the get-go.  There are no x and y that satisfy this equation, therefore there are no p and q that could exist with these properties, so to try to settle questions about whether or not "a" and "b" could be found is, well, I can see how it could be irritating, to say the least.

Many proofs of this type rely on the "coprimeness" of pairs of expressions.  During the course of the proof, it is quickly said that this expression and that one are coprime without giving a complete rationale.  This is done because it's aesthetically unpleasant to interrupt the flow of the proof to solve exercises that are regarded as "elementary".  One such exercise is: If gcd(2p, p2+3q2)=3, then given 3k=p, it follows that gcd(2k, 3k2+q2)=1.  The general case is:

Whenever gcd(a,b)=c, and c is not zero, you can validly conclude that gcd(a/c, b/c)=1

Proof: go back to the definition of gcd.  Three things are true: c|a, c|b, and if d|a and d|b then d|c.
Now, go back to the definition of "divides".  k and m exist such that ck=a and cm=b
Let x=gcd(k,m).  That means x|k, and x|m.
That means p and q exist such that xp=k and xq=m.  Plugging these into the equations of line 2 of this little proof,
cxp=a, and cxq=b, which by the definition of "divides" means that cx|a and cx|b.
By the "third true thing" of line 1 of this proof, cx|c, which means an integer, z, exists such that cxz=c
Therefore, x must be 1, so gcd(k,m)=1, so gcd(a/c,b/c)=1

Trust me, the proof doesn't need to be this long winded, and in fact, the statement is usually left without proof, because it is so elementary.  I just include it here to show you, in case you're not familiar with proofs involving gcd, how such a thing is proved.

Another one that comes up quite often is

Whenever gcd(a,b)=1 and gcd(d,b)=1, it follows that gcd(ad,b)=1.

Proof: suppose gcd(ad,b)=c, and c has a prime factor, p.
Then p|b, and p|ad, so either p|a or p|d
If p|a, then p|gcd(a,b), but no prime number divides 1, so p does not divide a.
If p|d, then p|gcd(d,b), but no prime number divides 1, so p does not divide d.
So, by contradiction, c has no prime factor.  Thus c must be 1.


Internet references

Euler�s Proof of the case n=3

Related Pages in this website

Complex Product Identity - (a2+b2)(A2+B2) = (aA+bB)2+(aB-bA)2 


The webmaster and author of this Math Help site is Graeme McRae.