``` We can repeat this operation which will lead to an infinite series of triangles with decreasing hypotenuses, thus not being possible. ``` I think the proof is not yet making this conclusion at this paragraph yet, it's merely generalizing the case where the sides have a common prime factor under the case where the sides are coprimes. Hence the following sentence: ``` By cancelling down as necessary, we can therefore assume the lengths to be coprime in pairs. ``` Euclid's formula is a fundamental formula for generating Pythagorean triples (three positive integers a, b, and c, such that $a^2 + b^2 = c^2$) given an arbitrary pair of integers $m$ and $n$ with $m > n > 0$. The formula states that the integers $$ a =m^{2}-n^{2} \\ b =2mn \\ c =m^{2}+n^{2} $$ form a Pythagorean triple. The triple generated by Euclid's formula is primitive - one in which $a, b$ and $c$ are coprime - if and only if $m$ and $n$ are coprime and not both odd. - If $y$ is a square then, $z=a^2+b^2=(\sqrt{z})^2$ which is just the Pythagorean theorem for the triangle on the right (Figure 3). - If $z$ is a square then, $y = a^2-b^2 \equiv a^2 = (\sqrt{y})^2+b^2 $ which is just the Pythagorean theorem for the triangle on the left (Figure 3). Note that one of $a$, $b$ has to be odd and the other one even. If $a$ is odd and $b$ is even, then $a^2$ is also odd and $b^2$ is even. Then both $a^2+b^2 = z$ and $a^2-b^2 = y$ are also odd, since odd +/- even = odd. The observation in the concluding paragraph is recognisable to students of mathematics. The method of ‘infinite descent’ is essentially a proof by induction, relying as it does on the well-foundedness of the natural numbers. Students will be familiar with the need to have a strong enough induction hypothesis to prove what is often a weaker statement. This corresponds to the ‘choice of conjecture’ of the article. If $y=n^2$ and $z=m^2$ it would be possible to construct a triangle with the length of the hypotenuse equal to $\frac{y+z}{2} = \frac{a^2+b^2+a^2-b^2}{2}= a ^2 = \frac{n^2 + m^2}{2} $ and the other sides equal to $\frac{z-y}{2}= \frac{a^2+b^2-a^2+b^2}{2}= b^2 = \frac{m^2 - n^2}{2}$ and $2\frac{m}{\sqrt{2}}\frac{n}{\sqrt{2}} = \sqrt{yz}$. Note that the new sides follow the structure defined by Euclid's formula. Pierre de Fermat was a French lawyer and mathematician born in 1607 who is given credit for his principle for light propagation and his Last Theorem. In number theory, Fermat's Last Theorem (FLT) states that no three positive integers $a, b$ and $c$ satisfy the equation $a^n + b^n = c^n$ for any integer value of $n$ greater than 2. Fermat claimed in 1637 in the margin of a copy of Arithmetica that he had a proof for FLT that was too large to fit in the margin. It took however 358 years of effort by some of the most talented mathematicians before Andrew Wiles released the first successful proof in 1994. In the meantime there were several proofs for specific exponents, including the one for $n=4$ which is particularly interesting because it was the only complete proof left by Pierre de Fermat.  Let's assume we have a triangle with sides $A^2$, $B^2$ and hypotenuse C. If 2 of the sides, for instance $A^2$ and $C$, have a common prime factor $p$, then $A^2=K_Ap$ and $C=K_Cp$ where $K_A,K_C$ are just the product of the rest of the prime factors. Using the Pythagorean Theorem $$ (A^2)^2+(B^2)^2=C^2 $$ and so $B^2=\sqrt{C^2-(A^2)^2}=\sqrt{K_C^2-K_A^2}p$ is also divisible by $p$. This way, it's always possible to divide all sides by $p$ getting a smaller triangle with integer lengths. We can repeat this operation which will lead to an infinite series of triangles with decreasing hypotenuses, thus not being possible.