L’échec de Terence Tao

We can first consider the fact that the roles of a and b are symetric. Moreover, if a or b is equal to 0, the affirmation is trivial and the same result stands if $a=b$. Therefore, let’s assume that a and b are non-zero positive integers. Then let’s use proof by contradiction: suppose that $(a_0,b_0)$ is a solution such that $\: \exists c \in \mathbb{N} \: \text{such that c is not a perfect square and} \: c=\frac{a_{0}^{2}+b_{0}^{2}}{a_{0}b_{0}+1}$ and where $a_{0} +b_{0}$ is minimal. WLOG, we can choose $a_0 \ge b_{0}$ ( because the roles of $a_{0}$ and $b_{0}$ are symetric and $a_{0}≠b_0$) : Thus, $a_{0}^{2}​−a_{0}.c^2.b_0​+(b_0^2​−c)=0$ This is a second-degree polynomial equation in $a_0​$ that has 2 solutions : $a_0​$ and $a$. By Vieta formula :

$$\left\{ \begin{array}{ll} a_{0} + a = c.b_{0} \\ a_{0}a = b_{0}^{2}-c \end{array} \right.$$Let’s show that $a < a_{0}$ and that $a \in \mathbb{Z}_{+}$:

First, $a=c.b_{0}-a_{0} \Rightarrow a\in \mathbb{Z}$
Then, $(a_{0} + a).a=a^{2}+b_{0}^{2}-c =a.b_0.c$.

However, $c < (a_{0}^{2} + b_0^2)$ because if not, by definition of $c$, then $a_0.b_0 \leq0$ which is false. So, $a = \frac{b_{0}^{2}+a_{0}^{2}-c}{b_{0}.c} > 0$.

Furthermore, since $a_{0} \neq 0$ and c is not a perfect square, $c \neq b_0^2 \Rightarrow a= \frac{b_{0}^{2}-c}{a_{0}} \neq 0$.
Moreover, by hypothesis and definition,$a_{0} \ge b_{0} \text{ and } (a_{0},b_{0},c) \in (\mathbb{N}^{*})^{3} \Rightarrow b_{0}^{2}- a_{0}^{2} \le 0<c \Rightarrow b_{0}^{2}-c<a_{0}^{2} \Rightarrow a.a_0<a_{0}^{2}\Rightarrow a<a_{0}$ because $a_0 \in \mathbb{N}^*$
Finally, $a \in \mathbb{Z}_+^{*} \text{ and } a<a_0$. As a consequence, $(a,b_{0})$ is another solution to the equation, but $a<a_0$, so $(a_0 + b_0)$ is not minimal : this is absurd. In conclusion, c is always a perfect square if c is an integer.