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Prove that $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$ is not an integer where $a$ and $b$ are positive integers.

One of the roots of the equation $10x^2+5x-1-(20a+1)b=0$ is $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$. Wolframalpha says that this equation has no integer solution. How can I prove that $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$ is not an integer?

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    $\begingroup$ Welcome to Math.SE! MathJax mode is started with $...$ for inline and $$...$$ for displayed equations. You should fix your expression there because the parentheses make no sense with the square root. $\endgroup$ – AlexR Mar 10 '15 at 12:04
  • $\begingroup$ Something is extremely wrong with that expression. The left part of the parenthesis appears inside the radical, while the right part of the parenthesis appears outside the radical. $\endgroup$ – barak manos Mar 10 '15 at 12:14
  • $\begingroup$ You can show that $b$ has to have the form $5n-1$, but I don't know if that's a useful step in proving the answer you want. $\endgroup$ – MonkeysUncle Mar 10 '15 at 12:49
  • $\begingroup$ If $\sqrt{5(160ab+8b+13)}$ is not integer, then so is the entire expression. Otherwise, it might help to notice: $5(160ab+8b+13)=20(40ab+2b+3)+5 \equiv 5\pmod{20}\implies\sqrt{5(160ab+8b+13)} \equiv 5,15 \pmod{20}$ $\endgroup$ – barak manos Mar 10 '15 at 12:58
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Take $b=14$ and $a=3$. In this case, $\sqrt{5(160ab+8b+13)} = 185$. This gives $x=\frac{185-5}{20}=9$ as a valid root. So there are integer solutions for $x$ when $a$ and $b$ are positive integers.

If Wolfram told you there were no integer solutions, either Wolfram is wrong, or perhaps you made a typo when entering your equation.

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