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- Thread starter jacks
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- Thread starter
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- Jan 26, 2012

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I'll get you started. What is an integer root? A quadratic equation [tex]x^2 + ax + b = 0[/tex] yields integer roots when [tex]\frac{-a \pm \sqrt{a^2 - 4b}}{2}[/tex] is an integer, with [tex]a, b \in \mathbb{Z}[/tex] ([tex]a[/tex] and [tex]b[/tex] don't need to be relative numbers but if we can prove there are infinitely many pairs of relative numbers [tex]\left (a, b \right )[/tex] we don't need to worry about non-integers).

Clearly if [tex]\frac{-a \pm \sqrt{a^2 - 4b}}{2}[/tex] is an integer, then [tex]-a \pm \sqrt{a^2 - 4b}[/tex] is one too.

Can you follow this reasoning to the end and establish a condition on [tex]a[/tex] and [tex]b[/tex] such that the resulting quadratic has integer roots? Then, do the same for the other equation with [tex]2a[/tex] instead and put all the conditions together. Then, prove that all these conditions are satisfied for infinitely many pairs [tex]\left (a, b \right )[/tex] and you will be done. Does that make sense?