<s>[INST] Does the equation have infinite number of solutions in integers? Does either of the equations ${a^2} - 10{b^2} =  \pm 1$ have infinite number of solutions in integers?
If the answers is yes, a hint about how to reduce this problem to the problem of Pythagorean triples in the Gaussian integers $\mathbb{Z}\left[ i \right]$ would be enough for solution.
If not, how would one proceed to prove that?
 [/INST] Clearly $$3^2 - 10 \cdot 1^2$$ is a solution.
Now observe that:


*

*$(a + b \sqrt{10})(a - b \sqrt{10}) = a^2 - 10 b^2$ for any integers $a,b$.

*$(a + b \sqrt{10})(c + d \sqrt{10}) = (ac+10bd) + (ad + bc)\sqrt{10}$


therefore every power $r>1$, $(3+\sqrt{10})^r$ will give a new solution of the Diophantine equation (clearly the components increase in size).



*

*$(3+\sqrt{10})^2 = (19 + 6 \sqrt{10})$ and $19^2 - 10\cdot 6^2 = 1$

*$(3+\sqrt{10})^3 = (117 + 37 \sqrt{10})$ and $117^2 - 10\cdot 37^2 = -1$

*...

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