Method of Infinite Descent

Proof by Infinite Descent

In this post we are going to go through a type of proof called Infinite Descent. Infinite Descent is a very effective method of solving Diophantine Equations.

Find all integer solutions to:

$$2a^2+b^2=2ab$$

Obviously, $(a,b)=0$ is a trivial solution, but can we find any non-trivial solutions?

Since $b$ must be even, let us define $b_1$ where $b=2b_1$.

So we have:

$$\begin{aligned} 2a^2+(2b_1)^2&=2a(2b_1) \\ 2a^2+4b_1^2&=4ab_1 \\ a^2+2b_1^2&=2ab_1 \end{aligned}$$

Since $a$ must be even, we define $a_1$ where $a=2a_1$.

$$\begin{aligned} (2a_1)^2+2b_1^2&=2(2a_1)b_1 \\ 4a_1^2+2b_1^2&=4a_1b_1 \\ 2a_1^2+b_1^2&=2a_1b_1 \end{aligned}$$

You can probably see where this is going. The equation above is just the original equation but $a_1$ and $b_1$ replaced $a$ and $b$.

Now, if we keep going, each $a_n$ can be replaced by a $a_{n+1}=\frac{a_n}2$.

Similarly, each $b_n$ can be replaced by a $b_{n+1}=\frac{b_n}2$.

If there is a solution $(a,b)$ that satisfies the original equation, the ordered pair $(a_1,b_1)=(\frac a2,\frac b2)$ is also a solution.

Now, since we are looking for integer solutions, Each $a$ and $b$ has a unique prime factorisation. All integers will have an integer number of factors of two.

Hence, at some point, $a_n$ and/or $b_n$ will not have a factor of $2$. This is a contradiction, since then the left hand side will be odd. Hence, there are no other solutions to $2a^2+b^2=2ab$ other than $(a,b)=(0,0)$.

The above example is an example of Infinite Descent.

Infinite Descent relies on the fact that there are a finite number of positive integers that are less than an integer. We first prove for something to be true for an integer, it must be also be true for a second smaller integer. This implying that a third smaller integer is also a solution. And a forth, a fifth, etc. Then by Mathematical Induction, there cannot exist any solutions, or else it would require a contradiction.