When the expression is a square of a rational number.

[chingel]

Homework Statement

I am trying to find when the square root of the expression [itex]25+8a^2[/itex] is rational, where the number a also needs to be rational. [itex]\sqrt{25+8a^2}=b[/itex], 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.

The Attempt at a Solution

I haven't really done a problem like this before, but it looks a lot like Pythagorean triples [itex]5^2+(2\sqrt{2}a)^2=c^2[/itex], so I tried working something with the formula [itex](m^2-n^2)^2+(2mn)^2=(m^2+n^2)[/itex]. I didn't have any other ideas than simply substituting something in there, for example [itex]2\sqrt{2}a=2mn[/itex]; then [itex]m=\sqrt{2}a/n[/itex]; then [itex]2a^2/n^2-n^2=5[/itex], 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.

 

[sankalpmittal]

Homework Statement

I am trying to find when the square root of the expression [itex]25+8a^2[/itex] is rational, where the number a also needs to be rational. [itex]\sqrt{25+8a^2}=b[/itex], 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.

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

If so then say ,

[itex]\sqrt{25+8a^2}[/itex] = p/q

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

Work it out.

 

[micromass]

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

[tex]x^2-8y^2 = 25[/tex]

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

 

[chingel]

Do you mean something like this?

[tex]a=\frac{5m}{2m^2-1}[/tex]
[tex]\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}}=[/tex][tex]\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}}=[/tex][tex]\frac{10m^2+5}{2m^2-1}[/tex]