# What are the necessary and sufficient conditions for a cubic equation to have integers roots

Let's start with Fermat equation with the lowest power, $x^3 + y^3 = z^3$. Now let's set $y = x + a, z = x + b$ with $b > a$ and $a,b$ integers. then the equation becomes

$$x^3 + (3a-3b)x^2 + (3a^2-3b^2)x + (a^3-b^3) = 0\tag 1$$

Since we know that Fermat's equation does not have a solution, we know that the above cubic will never admit integer solutions. The question is how to prove it.

• The way to prove (1) has no solutions is to note that it's equivalent to $x^3+y^3=z^3$ and then to use one of the proofs that that equation has no solutions. That's proved in lots of intro Number Theory textbooks, and no doubt there are many proofs on the web as well. – Gerry Myerson Nov 26 '14 at 5:43
• Sure but we learn nothing new by using the x^3 + y^3 = z^3 form of the equation. But we may learn something new by using the form above if someone can prove it does not have solutions in integers. – user25406 Nov 26 '14 at 17:16
• I think it is not realistic to expect this change of variables is going to lead to a genuinely new proof of Fermat's last theorem for exponent 3. I have never seen a proof of FLT for exponent 3 that did not involve a serious idea going beyond mere algebraic manipulations (e.g., passage to a more general ring of integers inside $\mathbf Q(\sqrt{-3})$). – KCd Dec 31 '14 at 17:32
• Well maybe it is not realistic but in mathematics we need proofs and sadly the words "not realistic" cannot be substituted for a proof that it cannot be done. – user25406 Jan 1 '15 at 13:45

Edit: complete restart. This is posted with the assumption that it is likely to have a fallacy, since it is too simple and obvious to have been missed for $350$ years, but with the hope that the methods presented may be useful.

Let $n=2m+1$ be an odd prime, assume that $x,y,z,a,b\in\Bbb Z$ and consider the solutions in $x$ of the following polynomial:

$$x^n+y^n=z^n=x^n+(x+a)^n=(x+b)^n\implies x^n+\sum_{i=0}^{n-1}{n\choose i}\left(a^{n-i}-b^{n-i}\right)x^i=0\tag 1$$

The solutions in $x$ become the symmetric polynomials of degree $1$ to $n$ of $n$ distinct solutions. Let these solutions be labeled $u_k, k\in[0,n-1]$ and let these polynomials take their values as follows:

$$\sum_{k=0}^{n-1}u_k = n(a-b)\tag {p_1}$$ $$\sum_{k=0,}^{n-2}\sum_{\ell=k+1}^{n-1}u_ku_\ell = {n\choose 2}(a^2-b^2)={n\choose 2}(a-b)(a+b)\tag {p_2}$$ $$\sum_{k=0,}^{n-3}\sum_{\ell=k+1,}^{n-2}\sum_{m=\ell+1}^{n-1}u_ku_\ell u_m = {n\choose 3}(a^3-b^3)={n\choose 3}(a-b)(a^2+ab+b^2)\tag {p_3}$$ $$\vdots$$ $$\prod_{k=0}^{n-1}u_k = a^n-b^n\tag{p_n}=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^i$$

Begin by claiming that at least one integer solution exists for $x$ in $(1)$; let this solution be the one labeled $u_0$ above. Each of the symmetric polynomials $(p_1),\dots,(p_n)$ above are integers on the RHS and must therefore be integers on the LHS as well. Since $u_0$ is claimed to be an integer, we must also have $\sum_{k=1}^{n-1}u_k$ be an integer in $(p_1)$. By the same token, $u_0\sum_{k=1}^{n-1}u_k$ is an integer in $(p_2)$ and therefore $\sum_{k=1,}^{n-2}\sum_{\ell=k+1}^{n-1}u_ku_\ell$ is also an integer in $(p_2)$. This process continues through the rest of these polynomials until we see that $\prod_{k=1}^{n-1}u_k$ must be an integer in both $(p_{n-1})$ and $(p_n)$.

Next, consider the divisibility facet of these polynomials. Each has a factor $a-b$ as demonstrated. Note this means that $a-b\mid x$ for any integer solution $x$. Each of the polynomials $(p_1),\dots,(p_{n-1})$ also has a factor $n$ since $n$ is prime and $\forall i\in[1,n-1],n\mid{n\choose i}$. Let $f$ be a prime factor of $a-b$. From $(p_n)$ we see that either $f\mid u_0$ or $f\mid\prod_{k=1}^{n-1}u_k$. In the symmetric polynomials $(p_1),\dots,(p_{n-1})$ there is more than one term and at least one term in each that does not involve $u_0$. Denote the portion of a given polynomial $(p_j)$ which contains only terms which have $u_0$ in them as $(p_j)[u_0]$. Then the portion of that polynomial which contains only terms which do not have $u_0$ in them is $(p_j)-(p_j)[u_0]$.

Now the process used to determine that $\forall j\in[1,n-1], (p_j)-(p_j)[u_0]\in\Bbb Z$ is repeated, except this time we are considering the given prime factor $f$. In particular, assume that $f\mid\prod_{k=1}^{n-1}u_k$. Then we have that $f\mid (p_{n-1})$ and also that $f\mid (p_{n-1})-(p_{n-1})[u_0]$ and therefore $f\mid (p_{n-1})[u_0] = u_0\sum_{k=1,}^{n-1}\prod_{\ell\neq k}u_\ell$. But $\sum_{k=1,}^{n-1}\prod_{\ell\neq k}u_\ell=(p_{n-2})-(p_{n-2})[u_0]$ and therefore we apply the same divisibility constraint to see that $f\mid (p_{n-2})[u_0]$. This process continues uninterrupted until we arrive at $f\mid u_0$ in $(p_1)$. We knew previously that either $f\mid u_0$ or $f\mid\prod_{k=1}^{n-1}u_k$; now we know that $f\mid u_0$, so we can start at $(p_1)$ and note that this means that $f\mid (p_1)-(p_1)[u_0]$, which further means that $f\mid (p_2)[u_0]$ and thus that $f\mid (p_2)-(p_2)[u_0]$, and again the process continues until we get back to $(p_n)$ and find that $f^2\mid (p_n)=(a^n-b^n)$. At this point, we either have $f^2\mid (a-b)$ or $f\mid\tfrac{a^n-b^n}{a-b}$. In the first case, we can jump to $(p_2)$ and note that $(p_2)=u_0((p_1)-(p_1)[u_0])+(p_2)-(p_2)[u_0]=n(a-b)(a+b)$ which means $f^2\mid (p_2), f^2\mid (p_2)[u_0]$ and therefore $f^2\mid (p_2)-(p_2)[u_0]$. But then this tracks back up to $(p_n)$ as $f^n\mid (p_n)$, and any further such factors demonstrate the same tendency, which leads to $C^n\mid a-b$ for some $C\in\Bbb Z$ unless we allow $f\mid \tfrac{a^n-b^n}{a-b}$. First, consider the branch assumption that $f\mid\tfrac{a^n-b^n}{a-b}$.

Consider $\tfrac{a^n-b^n}{a-b}=\sum_{i=0}^{n-1}a^{n-1-i}b^i\pmod{a-b}$. We have $f\mid a-b$ and $f\mid\tfrac{a^n-b^n}{a-b}$ so it is also the case that $f\mid\tfrac{a^n-b^n}{a-b}\pmod{a-b}$. It is relatively straightforward to show that $\sum_{i=0}^{n-1}a^{n-1-i}b^i\equiv na^{n-1}\equiv nb^{n-1}\pmod{a-b}$, therefore if $(f,n)=1$ then $f\mid a, f\mid b$. If $(f,n)=n$, then we go back to the symmetric polynomials, noting that while we haven't claimed $f^2\mid a-b$, we still have $\forall j\in[1,n], f^2\mid (p_j)$. In particular, we have $f^2\mid (p_{n-1})[u_0]$ and also $f^2\mid (p_{n-1})-(p_{n-1})[u_0]$ and $f\mid u_0$ and therefore $f^3\mid (p_n)$. But now we don't know whether $f^2\mid a-b$, and so we cannot yet extend this. We do know that $f^2\mid (a-b)^2$, so now consider

$$\sum_{i=0}^{n-1}a^{n-1-i}b^i\pmod{(a-b)^2}$$

With this particular modular equation, we cannot resolve it down to just one term on one side or the other immediately; instead, we have to systematically remove multiples of $a^2-2ab+b^2$ from both ends and meet in the middle. The first subtraction results in

$$3a^{n-2}b+\sum_{i=3}^{n-4}a^{n-1-i}b^i+3ab^{n-2}\equiv \sum_{i=0}^{n-1}a^{n-1-i}b^i\pmod{(a-b)^2}$$

while the second and further ones are similar,

$$\equiv 6a^{n-3}b^2-2a^{n-4}b^3+\sum_{i=4}^{n-5}a^{n-1-i}b^i-2a^{n-4}b^3+6a^{n-3}b^2\\ \equiv 10a^{n-4}b^3-5a^{n-5}b^4+\sum_{i=5}^{n-6}a^{n-1-i}b^i-5a^{n-5}b^4+10a^{n-4}b^3\\ \vdots\\ \equiv {m+2\choose 2}a^{m+1}b^{m-1}-\left(2\binom {m+1}2-1\right)a^mb^m+{m+2\choose 2}a^{m-1}b^{m+1}$$

Subtracting ${m+2\choose 2}a^{m-1}b^{m-1}(a-b)^2$ from this last expression yields $(2{m+2\choose 2}-2{m+1\choose 2}+1)(a^mb^m)$, and we see that $2{m+2\choose 2}-2{m+1\choose 2}+1=2m+1=n$, so now we have that $\tfrac{a^n-b^n}{a-b}\equiv na^{\tfrac {n-1}2}b^{\tfrac{n-1}2}\pmod {a^2-2ab-b^2}$, and therefore $f^2\mid na^mb^m$ and thus $f\mid ab$ ($f$ is prime). Since $f\mid ab$, now consider $\tfrac{a^n-b^n}{a-b}\equiv a^{n-1}+b^{n-1}\pmod {ab}$, which means that we have $f\mid a^{n-1}+b^{n-1}$. If we then subtract $a^{n-2}(a-b)$ from this, we get that $f\mid b^{n-1}\implies f\mid b\implies f\mid a$.

Since we have that $a-b\mid x$ and $f\mid a-b$ and now $f\mid a$ and $f\mid b$, assume that we have cleared away all such common factors, so $|f|=1$, which means that $a-b = -1$. The other possibility is that $f=a-b=0$, but this means that $x=0$ and $y=z$ which is a trivial solution. Since $f=1=1^n$, we can now rejoin our branch assumption to the other branch, namely that $f^n\mid a-b$.

This being done, our original equation looks like $x^n+y^n=(y+C^n)^n$. Now consider this equation from the perspective of $y$:

$$(y-a)^n+y^n=(y+c)^n\implies y^n+\sum_{i=0}^{n-1}{n\choose i}\left((-1)^{i+1}a^{n-i}-c^{n-i}\right)y^i=0$$

This equation is slightly different with the negative coefficients, so we create new symmetric polynomials for it with $v_k,k\in[0,n-1]$ the set of solutions and $(q_j),j\in[1,n]$ the set of polynomials. These polynomials and solutions can be defined simply, as before, and we will use $v_0$ as an intended integer solution. Rather than writing out the polynomials, we simply note important features of the set as a whole and only write out specific polynomials as needed. The first of these important features is to note that in this case, $c+a$ is a common factor to all; note that this means $c+a\mid y$. With $(q_2),(q_4),\dots,(q_{n-1})$ we have $(q_{2j})=c^{2j}-a^{2j}$ which has a factor $c^2-a^2=(c-a)(c+a)$. With $(q_1),(q_3),\dots,(q_n)$ we have $(q_{2j-1})=-c^{2j-1}-a^{2j-1}$ which has a factor $c+a$. The other important feature is that $\forall j\in[1,n-1],n\mid (q_j)$. But now we have defined the exact same scenario as before, with $c+a=b$ in the place of $a-b=c$. Finally, it is time to consider the equation from the perspective of $z$:

$$(z-b)^n+(z-c)^n=z^n$$

But wait, we have from before that $c=b-a\mid x$, and similarly $b=a+c\mid y$, which means that $\exists r,s\in\Bbb Z:rc=z-b, sb=z-c$. Rearranging, we get $(r-1)c=z-b-c=(s-1)b$ which means either $(b,c)\gt 1$ (contradicting the assumption that $(x,y,z)=1$) or else $z-b-c=bc\implies z=bc+b+c$. Since the former leads to a contradiction, we take $z=bc+b+c$ which means that we have completely specified all solutions in two parameters:

$$(bc+c)^n+(bc+b)^n=c^n(b+1)^n+b^n(c+1)^n=(bc+b+c)^n\tag 2$$

Consider the specific terms from each side which have exponent $1$ on either the $b$ or $c$ portion of the term. In particular, we see $n(bc)b^{n-1},n(bc)c^{n-1}$ appearing equally on both sides, while $nbc^{n-1}, nb^{n-1}c$ appear only on the RHS. In all other terms on both sides, there is a minimum factor $b^2c^2$. Therefore, $b^2\mid nbc,c^2\mid nbc$. We can resolve one or the other division by (w.l.o.g.) $c=1$, since we do not want $(b,c) \gt 1$. Going back to $(2)$,

$$(b+1)^n+2^nb^n=(2b+1)^n\\ \sum_{i=1}^{n}{n\choose i}b^i=\sum_{i=1}^{n-1}{n\choose i}2^ib^i\\ b^n=\sum_{i=1}^{n-1}{n\choose i}b^i(2^i-1)$$

If we divide away $b$ from both sides, we get $b^{n-1}=nbs+n$. Now we see that $b\equiv 0\pmod n$, but if we divide away $n$ from both sides, we get $b^{n-2}t=bs+1$. As $n\ge 3$, this last equation has no possible integer solutions, therefore one of our previous steps is the culprit. In particular, dividing by $b$ must have caused this issue, which would be a problem only if $b=0$. This is a valid solution, and as no other options are available, $b=0,c=1$ must be the entire set of solutions. But this means that $xyz=0$ is the only option for solutions to our original.

• Do you think the method above can be generalized to the general case of x^n + y^n = z^n and provide a way to prove Fermat theorem in a classical fashion ( not relying on modern mathematical theories). One obvious problem is the complexity of the algebra involved in dealing with the general case. – user25406 Nov 15 '14 at 13:20
• you would think that by now we would know everything we need to know about the roots of a cubic, their nature and their relation to the coefficients. – user25406 Nov 19 '14 at 12:44
• your equation 1, isn't it missing the variable x under the expansion? – user25406 Nov 21 '14 at 14:44
• Yes it is, good catch. I found another problem where $f^\infty$ is not the result, but rather $f^{tn}$, but I have a fix for that and I will update both as soon as I have some time. – abiessu Nov 21 '14 at 16:19
• And the same thing for the last equation in y^n ( also missing the y variable under the expansion). – user25406 Nov 21 '14 at 17:45