proposed elementary proof of Fermat’s last theorem

Michael Pogorsky has offered what is said to be an elementary proof of Fermat’s last theorem  . Is the proof correct? The intent of this entry is to show the proof up to the point at which it fails, if there is such a point. New equation numbers will be used.

proof We assume that there are positive integers $a,b$ and $c$ such that

 $a^{n}+b^{n}=c^{n}\quad\quad\quad(1)$

.

We can assume without loss of generality that $a,b$ and $c$ are mutually coprime, so that in fact they are also pairwise coprime. The proof is split into 3 major cases: (1) $n$ is a prime greater than 2, (2) $n$ is divisible by a prime greater than 2, and (3) $n$ is a power of 2.

1 n is a prime greater than 2

Write $c$ as

 $c=a+k=b+f$

for some integers $k$ and $f$.

Then

 $a^{n}+b^{n}=(a+k)^{n}=(b+f)^{n}.$

Using the binomial theorem we can write

 $a^{n}=f(nb^{n-1}+\frac{1}{2}n(n-1)b^{n-2}f+\cdots+f^{n-1})\quad\quad\quad(2)$

and

 $b^{n}=k(na^{n-1}+\frac{1}{2}n(n-1)a^{n-2}k+\cdots+k^{n-1}).\quad\quad\quad(3)$

Lemma 1. If $n$ is prime number  then $n$ divides $\binom{n}{k}$ for $0.
The proof is easy.

Claim: $gcd(f,k)=1$.
Proof. A factor of $f$ and $k$ will also divide $a$ and $b$ by equations (2) and (3). But $a$ and $b$ are coprime  , so the gcd of $f$ and $k$ must be 1.

Now write (2) as $a^{n}=fs$ for some integer $s$.

From that point in the proof the exposition is somewhat unclear so I will attempt to rearrange the steps in what seems to be a better order. First, I introduce a lemma of my own. Write (3) as $b^{n}=kt$ for some integer $t$.

Lemma 2. $gcd(f,s)=n^{\alpha}$ and $gcd(k,t)=n^{\beta}$ for some nonnegative integers $\alpha$ and $\beta$.
Proof. Suppose $q$ divides $\gcd(f,s)$ where $q$ is a prime. Then $q$ divides $a$ and $s$. We can write $s=nb^{n-1}+fT$ for some integer $T$, so that $q$ divides $s-fT$ and therefore $q$ divides $nb^{n-1}$. Hence $q$ divides $n$ or $q$ divides $b^{n-1}$. But if $q$ divides $b^{n-1}$ then $q$ divides $b$, a contradiction   . Hence $q$ divides $n$. But $n$ is a prime, so $q=n$. From this we get that $gcd(f,s)=n^{\alpha}$ for some nonnegative integer $\alpha$. Similarly, $gcd(k,t)=n^{\beta}$ for some nonnegative integer $\beta$.

It is clear that at least one of $\alpha$ and $\beta$ is zero, otherwise $n$ divides $f$ and $k$. Without loss of generality, we can assume that $\alpha=0$.

The author now introduces what he calls version A and version B. I would prefer to call these Case A and Case B. But there is no claim outstanding yet, so I have to defer the case split. What seems to be the next main result is stated in the following lemma.

Lemma 3. There exist positive integers $p,u,v,w$ such
1) $a=vp$,
2) if $\beta=0$ then

 $a=uwv+v^{n}$
 $b=uwv+w^{n}$
 $c=uwv+v^{n}+w^{n}.$

and
3) if $\beta>0$ then there is a positive integers $g$ such that

 $a=n^{g}uwv+v^{n}$
 $b=n^{g}uwv+n^{gn-1}w^{n}$

and

 $c=n^{g}uwv+v^{n}+n^{gn-1}w^{n}.$

Proof. (1) We have $a^{n}=fs$,where $f$ and $s$ are coprime. By unique factorization of integers it must be that $f=v^{n}$ and $s=p^{n}$ for some positive integers $p$ and $v$. It follows that $a=pv$.
(2) Similarly, there are positive integers $w$ and $q$ such that $b=wq$, where $w^{n}=k$. From $a+k=b+f$ we get

 $vp+w^{n}=wq+v^{n}$

and after regrouping we have

 $v(p-v^{n-1})=w(q-w^{n-1})$

Since $gcd(f,k)=1$ it follows that $gcd(v,w)=1$, so that

 $v|(q-w^{n-1})$

and

 $w|(p-v^{n-1}).$

Hence

 $u:=\frac{p-v^{n-1}}{w}=\frac{q-w^{n-1}}{v}\quad\quad\quad(4)$

is an integer. Using $u$ we can now write $a=uwv+v^{n}$, $b=uwv+w^{n}$, and $c=uwv+v^{n}+w^{n}$.
(3) Since $\alpha=0$, we have $gcd(f,s)=1$. Since $\beta>0$, we can write

 $k=k_{1}n^{\tau}$

where $\tau>0$ and $gcd(k_{1},n)=1$. By Lemma 1 there is a positive integer $c_{i}$ such that $\binom{n}{i}=nc_{i}$ for $0. We can write

 $t=\sum_{i=1}^{n}\binom{n}{i}a^{n-i}k^{i-1}=\sum_{i=1}^{n-1}nc_{i}a^{n-i}k^{i-1% }+k^{n-1}$
 $=\sum_{i=1}^{n-1}nc_{i}a^{n-i}k^{i-1}+nk_{1}n^{\tau-1}k^{n-2}=nT$

where $T=a^{n-1}+\frac{1}{2}(n-1)a^{n-2}k+\cdots+k_{1}n^{\tau-1}k^{n-2}$. Hence,

 $b^{n}=kt=knT=k_{1}n^{\tau+1}T.$

Claim: $gcd(T,n)=1$.
This follows from the fact that $n$ divides all the terms of $T$ except the first term. The first term is not divisible by $n$ because $k$ divides $n$ and therefore $n$ divides $b$ and $a$ and $b$ are coprime.
Claim: $gcd(T,k_{1})=1$.
This follows from the fact that $k$ divides $b$, so $k_{1}$ divides $b$, and $a$ and $b$ are coprime. By unique factorization of integers, then, it must be that there are positive integers $q$, $w$ and $\lambda$ such that $T=q^{n}$, $k_{1}=w^{n}$ and $n^{\tau+1}=\lambda^{n}$. Since $n$ is a prime, it follows that $\lambda=n^{g}$ for some positive integer $g$. Hence $gn=\tau+1$.

It follows that $b^{n}=w^{n}n^{gn}q^{n}$, so that $b=n^{g}wq$. From $a+k=b+f$ we get

 $vp+n^{gn-1}w^{n}=n^{g}wq+v^{n}$

which we can regroup to get

 $v(p-v^{n-1})=n^{g}w(q-n^{g(n-1)-1}w^{n-1}).$

Since $a$ and $b$ are coprime, it follows that $v$ and $n^{g}w$ are coprime. Hence

 $u:=\frac{p-v^{n-1}}{n^{g}w}=\frac{q-n^{g(n-1)-1}w^{n-1}}{v}$

is an integer. It follows that

 $vp-v^{n}=n^{g}wq-n^{gn-1}w^{n}=n^{g}uwv$

and one can now express $a$,$b$ and $c$ in terms of $u$,$v$,$w$.

Lemma 4. Let $u$ be the integer of Lemma 3. There is a monic polynomial  $P$ with integer coefficients  such that
(a) $P(u)=0$,
(b) the sum of the roots of $P$ is 0, and
(c) all coefficients of $P$ are divisible by $n$ except that when $\beta$ is positive the last coefficient is not divisible by $n$.
Proof. We use the same cases as in Lemma 3. (1) In this case we have

 $(uwv+v^{n})^{n}+(uwv+w^{n})^{n}-(uwv+v^{n}+w^{n})^{n}=0.$

The left hand side can be expanded using the binomial theorem to get a polynomial   $Q$ with coefficients that depend on $v$ and $w$. For the coefficient of $u^{n-i}$ we have to combine

 $\binom{n}{i}(uwv)^{n-i}(v^{n})^{i}+\binom{n}{i}(uwv)^{n-i}(w^{n})^{i}-\binom{n% }{i}(uwv)^{n-i}(v^{n}+w^{n})^{i}$
 $=u^{n-i}\binom{n}{i}(wv)^{n-i}(-(v^{n}+w^{n})^{i}+(v^{n})^{i}+(w^{n})^{i})% \quad\quad\quad(5)$

Clearly if $i=1$ this coefficient is 0. If $i=0$ the coefficient is $(wv)^{n}$. For the other terms we can write them as

 $\binom{n}{i}(wv)^{n-i}v^{n}w^{n}\sum_{j=1}^{i-1}\binom{i}{j}v^{n(j-1)}w^{n(i-j% -1)}$

so that the coefficient is divisible by $(wv)^{n}$. The coefficient is also divisible by $n$ if $1\leq i\leq n$ by Lemma 1. So we set $P:=Q/(wv)^{n}$ to get the conclusion  for case (1).
(2) We proceed as in case (1). The left side of the equation

 $(n^{g}uwv+v^{n})^{n}+(n^{g}uwv+n^{gn-1}w^{n})^{n}-(n^{g}uwv+v^{n}+n^{gn-1}w^{n% })^{n}=0$

can be expanded by the binomial theorem to get a polynomial $Q$ with coefficients that depend on $v$ and $w$. It is clear that the leading term is $(n^{g}uwv)^{n}$. For the coefficient of $u^{n-i}$ we have to combine

 $\binom{n}{i}(n^{g}uwv)^{n-i}(v^{n})^{i}+\binom{n}{i}(n^{g}uwv)^{n-i}(n^{gn-1}w% ^{n})^{i}-\binom{n}{i}(n^{g}uwv)^{n-i}(v^{n}+n^{gn-1}w^{n})^{i}$
 $=u^{n-i}\binom{n}{i}(n^{g}wv)^{n-i}((v^{n})^{i}+(n^{gn-1}w^{n})^{i}-(v^{n}+n^{% gn-1}w^{n})^{i}.\quad\quad\quad(6)$

This form makes it clear that the coefficient is 0 if $i=1$ and divisible by $n$ if $1\leq i. If $i=0$, the coefficient is $(n^{g}wv)^{n}$. Equation (6) is equal to

 $u^{n-i}\binom{n}{i}(n^{g}wv)^{n-i}\sum_{j=1}^{i-1}v^{n(j-1)}(n^{gn-1}w^{n})^{i% -j-1}v^{n}n^{gn-1}w^{n}$

which shows that $n^{gn}w^{n}v^{n}$ divides each coefficient. Set $P:=Q/(n^{g}wv)^{n}$ to get the conclusion for case (2).

Lemma 5. The polynomial $P$ of Lemma 4 has exactly one positive root.
Proof. By (5) and (6) the coefficients of $P$ are negative except for the leading coefficient. So there is exactly one sign change and by Descartes’s rule of signs there is exactly one positive root.

Definition. For each real root $u_{i}$ of $P$ we can define $a$, $b$ and $c$. (For example, $a=u_{i}wv+v^{n}$ and so on.) We say that a root $u_{i}$ is acceptable if the resulting $a,b,c$ are all positive integers.

Lemma 6. The only acceptable root of $P$ is $u$ and $u>0.$
Proof. Suppose that $u_{i}$ is a nonpositive acceptable root. Then $a,b,c$ are all positive and in case (1) we have

 $a+b=2u_{i}wv+v^{n}+w^{n}=c+u_{i}wv\leq c,$

while in case (2) we have

 $a+b=2n^{g}u_{i}wv+v^{n}+n^{gn-1}w^{n}=c+n^{g}u_{i}wv\leq c.$

But

 $a^{n}+b^{n}<(a+b)^{n}\leq c^{n}$

which is a contradiction. Since $u$ is acceptable, it must be that $u>0$.

The following lemma 7 is incorrect.
Lemma 7. $n$ does not divide $a+b$.
Proof. We use the cases of Lemma 3. (1) We write

 $a^{n}+b^{n}=(a+b)Q$

where

 $Q=\sum_{j=1}^{n}a^{n-j}b^{j-1}(-1)^{j-1}.$

It is known that the common divisor   if $a+b$ and $Q$ is $n$ and that if $n^{s}||a+b$ then $n^{s+1}||a^{n}+b^{n}$. Hence, we can write

 $a+b=n^{s}\delta$

and

 $Q=n\gamma$

where $gcd(n,\delta)=1$, $gcd(n,\gamma)=1$ and $gcd(\delta,\gamma)=1$. From $c^{n}=a^{n}+b^{n}=(a+b)Q=n^{s+1}\delta\gamma$ we get $s=n-1$, $n||c$ and

 $a+b=n^{n-1}\delta.\quad\quad\quad(7)$

Since $n$ divides $a+b$ and $c$ we have $n$ divides $2c-(a+b)=v^{n}+w^{n}$. It is also known that

 $v^{n}+w^{n}=(v+w)[(v+w)^{n-1}-nvw(v^{n-1}+\cdots+w^{n-1})]\quad\quad\quad(8)$

so that $n$ divides $v+w$. But then from (8) again, $n^{2}$ divides $v^{n}+w^{n}$. Now from (7) we have $n^{2}$ divides $a+b$. Hence

 $2c=(a+b)+v^{n}+w^{n}$

is divisible by $n^{2}$ and this is a contradiction.
(2) In this case

 $a+b=2n^{g}uwv+v^{n}+n^{gn-1}w^{n}$

so that if $n$ divides $a+b$ then $n$ divides $v^{n}$ and therefore $n$ divides $v$. From $a=vp$ we get then $n$ divides $a$ and therefore $n$ divides $b=a+b-a$. But $a$ and $b$ are coprime, so we have a contradiction.

Since Lemma 7 is incorrect, Lemma 8 is also incorrect.
Lemma 8. There are positive integers $u_{p}$ and $c_{p}$ such that $a+b={u_{p}}^{n}$ and $c=u_{p}c_{p}$.
Proof. We can write

 $x^{z}+y^{z}=(x+y)Q$

where

 $Q=\sum_{j=1}^{z}x^{z-j}y^{j-1}(-1)^{j-1}.$

It is an old result first attributed to Nicolas Malebranche (1638-1715) that if $x$ and $y$ are coprime and $d$ is a prime divisor   of $x+y$ and $Q$ then $d$ divides $z$. I will give a proof of this here. Define

 $Q_{1}=Q=x^{z-1}+x^{z-2}y+\sum_{j=3}^{z}x^{z-j}y^{j-1}(-1)^{j-1}.$

Let

 $Q_{2}=Q_{1}-x^{z-2}(x+y)=-2x^{z-2}y+x^{z-3}y^{2}+\sum_{j=4}^{z}x^{z-j}y^{j-1}(% -1)^{j-1}.$

Then $d$ divides $Q_{2}$. Define

 $Q_{3}=Q_{2}+2x^{z-3}y(x+y)=3x^{z-3}y^{2}-x^{z-4}y^{3}+\sum_{j=5}^{z}x^{z-j}y^{% j-1}(-1)^{j-1}.$

In general

 $Q_{n}=(-1)^{n+1}nx^{z-n}y^{n-1}+x^{z-n-1}y^{n}+\sum_{j=n+2}^{z}x^{z-j}y^{j-1}(% -1)^{j-1}$

and each $Q_{n}$ is divisible by $d$. Hence,

 $Q_{z}=\pm zy^{z-1}$

will be divisible by $d$. But $d$ does not divide $y$ (since otherwise it would also divide $x$, which would contradict that $x$ and $y$ are coprime). Hence, $d$ divides $z$. Using this result, we can say that if

 $a^{n}+b^{n}=(a+b)(a^{n-1}-a^{n-2}b+\cdots-ab^{n-2}+b^{n-1})=(a+b)Q$

then $a+b$ and $Q$ are coprime. Because if $d$ is a prime divisor of each then $d$ divides $n$, so $d=n$ and then $n$ divides $a+b$, which contradicts Lemma 7. By unique factorization there are positive integers $u_{p}$ and $c_{p}$ such that

 $a+b={u_{p}}^{n}$

and

 $Q={c_{p}}^{n}.$

Then $c^{n}=(a+b)Q={u_{p}}^{n}{c_{p}}^{n}$. Hence $c=u_{p}c_{p}$.

Lemma 9. Let $p,u,v,w$ be as in Lemma 3. Let $c_{p}$ be as in Lemma 8. Suppose there are positive integers $h$and $q$ such that

 $ah+bq=cp.$

Then one of the following possibilities holds:
(a) $h=h_{k}c,q=q_{k}c$ for some integers $h_{k}$ and $q_{k}$;
(b) $h=q=jc_{p}$ for some integer $j$;
(c) $h=jw^{n(n-1)},q=jv^{n(n-1)}$ for some integer $j$;
(d) $h=jb,q=jw^{n}$ for some integer $j$;
(e) $h=jv^{n},q=-jw^{n}$ for some integer $j$;
(f) $h=jv^{n},q=j(2uwv+w^{n}+2v^{n})$ for some integer $j$;
(g) $h=j(2uwv+2w^{n}+v^{n}),q=jw^{n}$ for some integer $j$.
Proof. At this point I think the proof is incomplete since he does not prove the result, but rather verifies that each of the possible solutions is indeed a solution. Later on,he needs to know that these are the only solutions.

2 n is divisible by a prime greater than 2

If $n=mz$ where $z$ is a prime greater than 2, then

 $(a^{m})^{z}+(b^{m})^{z}=(c^{m})^{z}$

and we can apply the results of section 1 to conclude that no such $z$ can exist.

3 n is a power of 2

It is known that if $n=4$ then Fermat’s Last Theorem is true. For example, see . So if $n=2^{t},t\geq 3$, then we can write

 $(a^{2^{t-2}})^{4}+(b^{2^{t-2}})^{4}=(c^{2^{t-2}})^{4}$

which contradicts the theorem for $n=4$.

References

• 1 G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford University Press, page 191.
Title proposed elementary proof of Fermat’s last theorem ProposedElementaryProofOfFermatsLastTheorem 2013-03-22 17:36:55 2013-03-22 17:36:55 Mathprof (13753) Mathprof (13753) 78 Mathprof (13753) Proof msc 11D41