Skip to main content

Section 1.2 The Diophantine equation \(a^2+b^2=c^2\)

In calculus classes the world over we use blindly that in a right triangle, that the square of the hypotenuse is equal to the sums of the square of the other two sides. Indeed by just looking at the first few integer solutions 1  to

\begin{equation} a^2+b^2=c^2\tag{1.2.1} \end{equation}

we can see a pattern.

\begin{align*} 3^2+4^2 \amp= 5^2\\ 8^2+6^2 \amp= 10^2\\ 15^2+8^2 \amp= 17^2\\ 24^2+10^2 \amp= 26^2\\ 35^2+12^2 \amp= 37^2\\ \vdots\\ (n^2-1)^2+(2n)^2 \amp= (n^2+1)^2. \end{align*}

If we consider \(n=\frac{p}{q}\) to be a rational number and clear denominators we get the general solution

\begin{equation*} (p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2. \end{equation*}

There is a wonderful proof of Pythagoras' theorem which we will consider now.

A Diophantine equation is a polynomial in which we only consider variables to be integers.

Without loss of generality, consider Figure 1.2.2 below. The area of the outside square is \(c^2.\) This is equal to the area of the inside square plus that of the four triangles inside the large square. Thus

\begin{align*} c^2 \amp=(b-a)^2+4\left(\frac{ab}{2}\right)\\ \amp=b^2-2ab+a^2+2ab\\ \amp=a^2+b^2 \end{align*}

which is the desired result, proving Pythagoras' theorem.

Figure 1.2.2. This picture is due to Khan Amore.

After considering the diophantine equation \(a^2+b^2=c^2,\) it is almost immediate to inquire about solutions to \(a^3+b^3=c^3,\) \(a^4+b^4=c^4,\) or in general \(a^k+b^k=c^k\) for \(k\gt 2.\) Unfortunately this is a very hard thing to consider. The question was posed by Pierre de Fermat in 1637, and remained unproven until 357 years later in 1995, Andrew Wiles succeeded in proving it. Wiles proved the following theorem.

This proof takes up around 200 pages, and so we will not reproduce it.