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?

  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.