Fermat’s proof of descent for n=4

The exponent-four case of Fermat’s Last Theorem says that there are no nonzero integers $x,y,z$ satisfying

$\displaystyle x^{4}+y^{4}=z^{4}.$

Fermat’s original method proves something stronger and, in a sense, more natural: $\displaystyle x^{4}+y^{4}=z^{2}$ has no solution in positive integers. Once this stronger statement is known, the exponent-four case follows at once, because a hypothetical equation $x^{4}+y^{4}=z^{4}$ would be an equation of the stronger kind with $z^{2}$ in place of the final square. The reason that the stronger statement is the right target is geometric. A relation $x^{4}+y^{4}=z^{2}$ says that $x^{2},y^{2},z$ are the side lengths of a right triangle: the two legs are themselves squares. Fermat’s descent exploits the impossibility of repeatedly extracting smaller and smaller such right triangles.

The proof is elementary in the sense that it stays entirely within ordinary integers. Unlike the exponent-three proof, there is no need to move to a triangular complex grid. But the underlying pattern is very similar. We begin with a hypothetical smallest solution, use a factorization to split it into coprime pieces, observe that each piece must itself be a square, and then reassemble those squares into a new solution with a smaller final term. The mathematics is not just a chain of lucky substitutions. Every substitution is forced by trying to expose the next factorization.

We will repeatedly use one simple principle. If two coprime positive integers have a product that is a square, then each of them is a square. To see this, factor both numbers into primes. Since the two numbers have no prime in common, every prime exponent appearing in either one is also the exponent of that prime in the product. If the product is a square, every such exponent is even, and hence each factor is itself a square. This elementary observation is the engine that will run the descent twice.

Assume, for contradiction, that there are positive integers $x,y,z$ with

$\displaystyle x^{4}+y^{4}=z^{2}.$

Choose such a solution with $z$ as small as possible. We may assume that $x$ and $y$ are coprime. Indeed, if a number $d>1$ divided both $x$ and $y$ , then $d^{4}$ would divide the left side and therefore divide $z^{2}$ . Looking prime by prime, this forces $d^{2}$ to divide $z$ , and then $(x/d)^{4}+(y/d)^{4}=(z/d^{2})^{2}$ would be a solution with a smaller final square root. This would contradict our choice of $z$ .

The two numbers $x$ and $y$ cannot both be odd. Every odd fourth power is congruent to $1$ modulo $16$ , so the sum of two odd fourth powers would be congruent to $2$ modulo $16$ . But the possible square residues modulo $16$ are only $0,1,4,$ and $9$ . Thus exactly one of $x,y$ is even. By interchanging them if necessary, let $x$ be even and $y$ odd. The equation now says that

$\displaystyle (x^{2})^{2}+(y^{2})^{2}=z^{2}.$

So $x^{2},y^{2},z$ form a primitive Pythagorean triple. Rather than merely quote the usual classification of such triples, let us derive precisely the piece of it that we need. Because $x$ is even and $y$ is odd, the number $z$ is odd. Rearranging the Pythagorean equation gives $x^{4}=z^{2}-y^{4}$ , and hence

$\displaystyle x^{4}=(z+y^{2})(z-y^{2}).$

Both factors on the right are positive and even, because $z$ and $y^{2}$ are odd. Write $z+y^{2}=2M$ and $z-y^{2}=2N$ , where $M>N>0$ . Then $(x^{2}/2)^{2}=MN$ . The numbers $M$ and $N$ are coprime. A common divisor of them would divide their sum $M+N=z$ and their difference $M-N=y^{2}$ ; but any prime dividing both $z$ and $y$ would also divide $x$ , contrary to the coprimality of $x$ and $y$ .

Since $M$ and $N$ are coprime and their product is a square, they must separately be squares. Write $M=m^{2}$ and $N=n^{2}$ , with $m>n>0$ . Adding and subtracting the defining equations for $M,N$ yields

$\displaystyle x^{2}=2mn,\quad y^{2}=m^{2}-n^{2},\quad z=m^{2}+n^{2}.$

This is the primitive Pythagorean-triple parametrization in the particular form we need. Notice how it arose. The first factorization showed that the large square $x^{4}$ had been divided into two coprime factors; therefore each factor was itself a square. The quantities $m,n$ are not guessed—they are the square roots forced by coprimality.

The numbers $m$ and $n$ are coprime and have opposite parity, because $m^{2}-n^{2}=y^{2}$ is odd. We must determine which one is even. If $m$ were even and $n$ odd, then $m^{2}-n^{2}$ would be congruent to $0-1$ , or $3$ , modulo $4$ . But no square is congruent to $3$ modulo $4$ . Therefore $m$ is odd and $n$ is even.

The equation $x^{2}=2mn$ now becomes much more informative. Since $n$ is even, write $n=2n_{0}$ . Then

$\displaystyle \left(\frac{x}{2}\right)^{2}=mn_{0}.$

The two factors $m$ and $n_{0}$ are coprime. Their product is a square. Hence each is a square. There are coprime positive integers $u,v$ such that $\displaystyle m=u^{2},\quad n=2v^{2}.$ The parity information tells us that $u$ is odd. Substitute this into the other equation $y^{2}=m^{2}-n^{2}$ . We obtain

$\displaystyle y^{2}=u^{4}-4v^{4}=(u^{2}-2v^{2})(u^{2}+2v^{2}).$

This is the second decisive factorization. The two factors are positive because $m>n$ , which says $u^{2}>2v^{2}$ . They are odd because $u$ is odd. They are also coprime. Indeed, any common divisor divides their sum $2u^{2}$ and their difference $4v^{2}$ . Since it is odd, it must divide both $u^{2}$ and $v^{2}$ ; but $u,v$ are coprime. Thus the common divisor is $1$ .

The product of the two coprime positive numbers $u^{2}-2v^{2}$ and $u^{2}+2v^{2}$ is the square $y^{2}$ . Each factor must therefore be a square. Let

$\displaystyle u^{2}+2v^{2}=P^{2},\quad u^{2}-2v^{2}=Q^{2},$

where $P>Q>0$ . Both $P$ and $Q$ are odd, because their squares are odd. The two equations have a simple geometric meaning: they say that $u^{2}$ lies exactly halfway between two odd squares, while the distance from it to either one is $2v^{2}$ . But their real force comes from subtracting them:

$\displaystyle P^{2}-Q^{2}=4v^{2}.$

Factor the difference of squares. Since $P,Q$ are odd, both $P+Q$ and $P-Q$ are even. Define $\displaystyle R=\frac{P+Q}{2},S=\frac{P-Q}{2}.$ Then $R$ and $S$ are positive integers, and the difference-of-squares identity says simply that $RS=v^{2}$ . They are coprime. Any common divisor of $R$ and $S$ divides $R+S=P$ and $R-S=Q$ . But $P$ and $Q$ are coprime, because their squares were the coprime numbers $u^{2}+2v^{2}$ and $u^{2}-2v^{2}$ .

Once again, a product of coprime integers is a square. Thus $R$ and $S$ must each be squares. Write $\displaystyle R=a^{2},S=b^{2}.$ The identities $P=R+S$ and $Q=R-S$ now become $P=a^{2}+b^{2}$ and $Q=a^{2}-b^{2}$ . Recall that $u^{2}$ is the average of $P^{2}$ and $Q^{2}$ , because the two earlier equations add to $2u^{2}=P^{2}+Q^{2}$ . Substituting the expressions for $P,Q$ , we find

$\displaystyle u^{2}=\frac{(a^{2}+b^{2})^{2}+(a^{2}-b^{2})^{2}}{2}=a^{4}+b^{4}.$

This is the new solution. From the hypothetical solution $x^{4}+y^{4}=z^{2}$ , we have constructed another solution $\displaystyle a^{4}+b^{4}=u^{2}.$

It remains only to check that it is genuinely smaller in the way required for descent. The old final term was $\displaystyle z=m^{2}+n^{2}=u^{4}+4v^{4}.$ Since $u,v$ are positive, $u<z$ . Thus $a^{4}+b^{4}=u^{2}$ is another positive solution of exactly the same kind, but with final square root $u$ strictly less than the supposedly minimal $z$ . This contradiction proves that no such initial solution existed.

Therefore $\displaystyle x^{4}+y^{4}=z^{2}$ has no solution in positive integers. In particular, if $x^{4}+y^{4}=z^{4}$ had a nonzero integer solution, then it would provide a solution of the forbidden stronger equation with $z^{2}$ in place of the final variable. Hence $\displaystyle x^{4}+y^{4}=z^{4}$ has no nonzero integer solutions.

The proof is worth pausing over because its internal architecture is so clean. One begins with a square that is a sum of two fourth powers. Since fourth powers are squares of squares, the equation is a Pythagorean-triangle equation whose legs are squares. The classification of primitive right triangles breaks those legs into coprime factors. The fact that the legs are themselves squares then forces the Pythagorean parameters to contain further squares. A second factorization produces another pair of coprime factors, which must again be squares. Finally, the algebra rearranges itself into a new equation of the original form.

The descent can be summarized as a chain of forced square roots:

$\displaystyle x^{4}+y^{4}=z^{2}\quad\Longrightarrow\quad x^{2}=2mn,\;y^{2}=m^{2}-n^{2}\quad\Longrightarrow\quad m=u^{2},\;n=2v^{2}$

and then

$\displaystyle y^{2}=(u^{2}-2v^{2})(u^{2}+2v^{2})\quad\Longrightarrow\quad u^{2}+2v^{2}=P^{2},\;u^{2}-2v^{2}=Q^{2}$

which finally yields $\displaystyle a^{4}+b^{4}=u^{2},\quad 0<u<z.$

The original solution contains, hidden inside it, a smaller solution. That is the essential paradox. The argument does not say that one literally performs infinitely many computations. It says that if any solution existed, there would be a smallest one; but the arithmetic of that smallest one manufactures a smaller one. The first assumed solution therefore cannot exist at all.

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