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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?

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you will have better luck reducing the problem to the problem of finding the group of units in $\Bbb Z[\sqrt{10}]$ –  mercio Nov 21 '12 at 14:30
    
Thanks, I'll try that now –  Alen Nov 21 '12 at 14:32
    
In addition, you could try to use the continued fractions. –  awllower Nov 21 '12 at 14:57

2 Answers 2

up vote 4 down vote accepted

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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1  
My thoughts excatly :D. Thank you –  Alen Nov 21 '12 at 14:40

Function $N:\mathbb{Q} + \mathbb{Q}\sqrt {10} \to \mathbb{Z}$ defined by $N\left( {a + b\sqrt {10} } \right) = {a^2} - 10{b^2}$ is multiplicative.

Also, it can be shown that the element of $\mathbb{Q} + \mathbb{Q}\sqrt {10} $ is a unit iff $N\left( {a + b\sqrt {10} } \right) = \pm 1$.

$N\left( {3 + \sqrt {10} } \right) = 9 - 10 = - 1$ so $N\left( {{{\left( {3 + \sqrt {10} } \right)}^n}} \right) = {\left( { - 1} \right)^n}$.

So, it has infinite number of solutions.

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