# When the expression is a square of a rational number.

1. Aug 7, 2012

### chingel

1. The problem statement, all variables and given/known data
I am trying to find when the square root of the expression $25+8a^2$ is rational, where the number a also needs to be rational. $\sqrt{25+8a^2}=b$, where a and b are both rational numbers. I am trying to get an expression for a in terms of some other number m, which would always make the number a satisfy the original requirement.

3. The attempt at a solution

I haven't really done a problem like this before, but it looks a lot like Pythagorean triples $5^2+(2\sqrt{2}a)^2=c^2$, so I tried working something with the formula $(m^2-n^2)^2+(2mn)^2=(m^2+n^2)$. I didn't have any other ideas than simply substituting something in there, for example $2\sqrt{2}a=2mn$; then $m=\sqrt{2}a/n$; then $2a^2/n^2-n^2=5$, but this just gets me back to the beginning, if I multiply by n^2 and try to solve using the quadratic formula.

Any hints are appreciated. Some examples of a's that work are 0, 10/7 and 15/17.

2. Aug 8, 2012

### sankalpmittal

Well , I don't quite understand what you are trying to prove. Are you trying to prove that if $\sqrt{25+8a^2}$ is rational , then a is also rational ?

If so then say ,

$\sqrt{25+8a^2}$ = p/q

where p and q are coprimes and q is not 0.

Work it out.

3. Aug 8, 2012

### micromass

Staff Emeritus
So basically, you need find the rational points on the hyperbola

$$x^2-8y^2 = 25$$

See this link: http://mathcircle.berkeley.edu/BMC4/Handouts/elliptic/node4.html [Broken] to find the rational points on the circle. Can you adapt the proof such that it works in this case?

Last edited by a moderator: May 6, 2017
4. Aug 8, 2012

### chingel

Do you mean something like this?

$$a=\frac{5m}{2m^2-1}$$
$$\sqrt{25+8a^2}=\sqrt{25+8\frac{25m^2}{(2m^2-1)^2}}=\sqrt{\frac{25(2m^2-1)^2+200m^2}{(2m^2-1)^2}}=$$$$\sqrt{\frac{100m^4-100m^2+25+200m^2}{(2m^2-1)^2}}=\sqrt{\frac{100m^4+100m^2+25}{(2m^2-1)^2}}=\sqrt{\frac{(10m^2+5)^2}{(2m^2-1)^2}}=$$$$\frac{10m^2+5}{2m^2-1}$$