# The Method of Infinite Descent – Fermat’s Favourite Proof

Fermat left only one proof.

The area of a Pythagorean triangle is never a square number.

Fermat wrote ,  “If the area of a right-angled triangle were a square, there would exist two biquadrates (fourth powers) the difference of which would be a square number.”

That is,

$$a^4\, -\, b^4 = c^2$$

He used the method of infinite descent which is a proof of non-existence or a form of proof by contradiction.  It was not a new proof – it is recorded in Euclid’s Elements.

It relies on the Well-Ordering Principle that a set of numbers contains a least element.

One assumes that a solution exists in the natural numbers which would imply that a second, smaller solution also exists and so on.  Since there cannot exist an infinity of of ever smaller solutions in the natural numbers, the original premise is incorrect and the assumption is contradicted.

The descent step seems to construct a set of positive integers which does not have a least member.  Therefore, it is the empty set.

We consider the related Diophantine equation

$$x^4 + y^4 = z^2$$

and whether it has any solutions.

Let the solution be :

$$x = x_1,\,\, y = y_1,\,\, z = z_1$$.

If gcd $$(x_1, y_1) = d \gt 1$$, then we can write $$x_1 = dx’,\, y_1 = dy’$$.

Substitute into the equation :

$$(dx’)^4 + (dy’)^4 = z_1^2$$

Factorise :

$$d^4(x’^4 + y’^4) = z_1^2$$

It follows that $$d^2$$ divides $$z_1$$ and that $$z_1 = d^2z’$$.

$$(z_1^2/d^4 = z’^2,\,\, z_1^2 = d^4z’^2,\, \,z_1 = d^2z’)$$

Then :

$$x’^4 + y’^4 = z’^2$$,

where gcd$$(x’, y’) = 1$$ and $$z’ \lt\, z$$.  We can assume that gcd $$(x_1, y_1) = 1$$.  We have established that they have no common factors.

We now write the equation as

$$(x_1^2)^2 + (y_1^2)^2 = z_1^2$$

This is a primitive Pythagorean triple.

\begin{align} x_1^2 &= m^2\, – \,n^2 \tag 1\\ y_1^2 &= 2mn \tag2\\ z_1 &= m^2 + n^2 \tag3 \end{align}

where $$m$$ and $$n$$ are relatively prime positive integers of opposite parity and $$m\gt n$$.

In this case, $$m$$ must be odd and $$n$$ must be even otherwise

$$x_1^2 = m^2\,-\,n^2\equiv 0\, – 1 \equiv 3 \pmod 4$$

which is impossible because a square number is congruent to either $$0$$ or $$1$$ in modulo $$4$$.

Let :

$$n = 2r$$

Modify the equations :

\begin{align} x_1^2 &= m^2 \,- \,4r^2\tag4\\ y_1^2 &= 4mr \tag5\\ z_1 &= m^2 + 4r^2\tag6 \end{align}

From ($$4$$) :

$$x_1^2 + 4r^2 = m^2$$

so $$(x_1, 2r, m)$$ is a primitive Pythagorean triple.

From $$(5)$$ :

$$(y_1/2)^2 = mr$$

where $$m$$ and $$r$$ are relatively prime; it follows that $$m$$ and $$r$$ are both squares.

Let :

$$m = s^2$$ and $$r = t^2$$

We descend.

\begin{align} x_1 &= u^2\,-\,v^2\tag7\\ 2r &= 2uv\tag8\\ m&= u^2 + v^2\tag9 \end{align}

where $$u$$ and $$v$$ are relatively prime positive integers with opposite parity and $$u\gt v$$

But :

$$n = 2r = 2t^2$$

So :

$$uv = t^2$$

Hence, $$u$$ and $$v$$ are both squares.

Let :

$$u = x_2^2,\,v = y_2^2$$

Substitute into $$(9)$$ above :

$$s^2 = x_2^4 + y_2^4 = z_2^2 \text\,{,say} \tag{10}$$

Thus, this is another solution to the original equation.

It is ‘smaller’ because

$$0 \lt z_2 = s \le m \lt m^2 + n^2 = z_1$$

We have completed the descent step.

Thus, the assumption that there exists a positive solution is contradicted.

The Diophantine equation

$$x^4 + y^4 = z^2$$

has no positive solutions.

Since any fourth power is necessarily a square, the method of infinite descent can be used to show that the Diophantine equation

$$x^4 +y^4 = z^4$$

has no positive solutions.