Are there infinitely many rational pairs $(a,b)$ which satisfy given equation?

Question

I saw on some facebook page this concrete example:

$1.2^2+0.6^2=1.2+0.6$

The question that immediately arises is:

Are there infinitely many pairs $(a,b)$ of rational numbers such that we have $a^2+b^2=a+b$?

Thank you for your response.

Answer 1

Hint: $$a^2+b^2=a+b\implies{}a^2-a+b^2-b=0\implies{}2\left(a-\frac{1}{2}\right)^2+2\left(b-\frac{1}{2}\right)^2=1.$$

What can you say about this equation with respect to the cartesian plane? What is the parametrization of this figure?

Answer 2

Real solutions:

The equation $$ a^2 + b^2 = a + b $$ can be transformed: $$ a^2 + b^2 = a + b \iff \\ a^2 - a + b^2 -b = 0 \iff \\ (a - 1/2)^2 + (b - 1/2)^2 = 1/4 + 1/4 = 1/2 = (1/\sqrt{2})^2 \quad (*) $$ The solutions form a circle with origin $(1/2, 1/2)$ and radius $1/\sqrt{2}$ in $\mathbb{R}^2$. They are $$ (a, b) = (a, b(a)) = \left(a, (1/2) \pm\sqrt{(1/\sqrt{2})^2 - (a - 1/2)^2}\right) $$ for $a \in [1/2 - 1/\sqrt{2}, 1/2 + 1/\sqrt{2}] = [a_-, a_+]$.

Rational solutions:

Among all those infinite many real solutions $(a, b)$ we expect infinite many rational solutions, but my topology is too bad to give a good argument here. If we insert arbitrary rational $a$ not all $b = b(a)$ are rational.

I tried to find all rational $a$ for which $b = b(a)$ is rational, but ran against a wall. However an infinite subset of such rational $a$ is still infinite, so we try to pick an easy one.

An infinite rational subset of solutions:

The subset I came up with consists of those $a$ which have the form $$ a_n = \frac{1}{2} + \frac{1}{2n} < a_+ \quad (n \in \mathbb{N}) $$ which gives \begin{align} b_n &= \frac{1}{2} \pm \sqrt{\frac{1}{2} - \frac{1}{4n^2}} \\ &= \frac{1}{2} \pm \frac{\sqrt{2n^2 - 1}}{2n} \\ &= \frac{n \pm \sqrt{2n^2 - 1}}{2n} \end{align}

Then $b_n$ is rational, if $$ 2n^2 -1 = m^2 $$ for some $m \in \mathbb{N} \cup \{ 0 \}$.

We can rewrite this as $$ m^2 - 2 n^2 = -1 \quad (**) $$ which is a Diophantine equation related to the Pell equation with constant $D = 2$, see negative Pell equation.

Solution of the negative Pell equation:

One solution to $(**)$ is $m = 1$ and $n = 1$, from which we can generate infinite many more solutions:

For odd $k \in \mathbb{N}$ we have: $$ -1 = (1^2 - 2\cdot 1^2)^k = (m^2 - 2n^2) \Rightarrow \\ -1 = (1 + \sqrt{2})^k (1-\sqrt{2})^k = (m + \sqrt{2} n)(m - \sqrt{2} n) $$ which because of $$ (1 + \sqrt{2})^k = \sum_{i=0}^k \binom{k}{i} 1^i (\sqrt{2})^{n-i} = c_1 \cdot 1 + c_2 \sqrt{2} \quad (c_1, c_2 \in \mathbb{N}) \\ (1 - \sqrt{2})^k = \sum_{i=0}^k \binom{k}{i} 1^i (-\sqrt{2})^{n-i} = d_1 \cdot 1 - d_2 \sqrt{2} \quad (d_1, d_2 \in \mathbb{N}) $$ gives $$ m + \sqrt{2} n = (1 + \sqrt{2})^k \\ m - \sqrt{2} n = (1 - \sqrt{2})^k \\ $$ and $$ m = \frac{(1 + \sqrt{2})^k + (1 - \sqrt{2})^k}{2} \\ n = \frac{(1 + \sqrt{2})^k - (1 - \sqrt{2})^k}{2 \sqrt{2}} $$

Here are the first ten solutions:

k = 1,  m = 1,       n = 1
k = 3,  m = 7,       n = 5
k = 5,  m = 41,      n = 29
k = 7,  m = 239,     n = 169
k = 9,  m = 1393,    n = 985
k = 11, m = 8119,    n = 5741
k = 13, m = 47321,   n = 33461
k = 15, m = 275807,  n = 195025
k = 17, m = 1607521, n = 1136689
k = 19, m = 9369319, n = 6625109

Plugging the Pell solutions into the circle solution:

We have \begin{align} a_k &= \frac{1}{2} + \frac{1}{2n} \\ &= \frac{1}{2} + \frac{\sqrt{2}}{(1 + \sqrt{2})^k - (1 - \sqrt{2})^k} \in \mathbb{Q} \end{align} and \begin{align} b_k &= \frac{1}{2} \pm \frac{m}{2n} \\ &= \frac{1}{2} \pm \frac{\sqrt{2}}{2} \frac{(1 + \sqrt{2})^k + (1 - \sqrt{2})^k} {(1 + \sqrt{2})^k - (1 - \sqrt{2})^k} \in \mathbb{Q} \end{align} for any odd $k \in \mathbb{N}$.

Here is a visualization:

(Large version)

Featured are the real solutions (blue circle) and a few of the rational solutions $Q_k^{(+)} = (a_k, b_k^{(+)})$ (green points), $Q_k^{(-)} = (a_k, b_k^{(-)})$ (red points) for $k \in \{ 1,3,5 \}$. The other ones pile up too much to be distinguishable in the plot.