4
$\begingroup$

Find naturals numbers $p$ and $q$ such that $\frac{\sqrt{p^2+7}}{\sqrt{q^2-3}}$ is rational

i have taken $$\frac{\sqrt{p^2+7}}{\sqrt{q^2-3}}=\frac{a}{b}$$ where $a$ and $b$ are positive integers which are co prime then after simplification we get

$$3a^2+7b^2=(aq-bp)(aq+bp)$$

$$3(a^2+b^2)+4b^2=(aq-bp)(aq+bp)$$

Obviously $a$ and $b$ both cannot be even since they are co prime

but now if i consider other cases i could not find $p$ and $q$...i will be happy if i get any hint

$\endgroup$
5
  • 4
    $\begingroup$ hint: don't make a system of equations. just consider the numerator and denominator separately; you want them both to be integers. $\endgroup$
    – hunter
    May 2, 2017 at 3:12
  • 1
    $\begingroup$ Ok fine got it $p=3$ and $q=2$ $\endgroup$ May 2, 2017 at 3:18
  • 1
    $\begingroup$ but what i thought is even though both are irrational ratio can be rational like $\frac{\sqrt{3}}{2\sqrt{3}}$ is a rational $\endgroup$ May 2, 2017 at 3:20
  • $\begingroup$ If $p^2 + 7 = a^2*k$ and $q^2 - 3 = b^2*k$ we are still solving for integers $\endgroup$
    – fleablood
    May 2, 2017 at 3:33
  • 2
    $\begingroup$ Here are all solutions $p$, $q$, $r=\displaystyle\frac{\sqrt{p^2+7}}{\sqrt{q^2-3}}$ for $p\in[1,10^6]$, $q\in[1,200]$: $$p=3,\quad q=2,\quad r= 4 $$ $$p=9,\quad q= 5,\quad r= 2 $$ $$p=27,\quad q= 7,\quad r= 4 $$ $$p=75,\quad q= 5,\quad r= 16 $$ $$p=3147,\quad q= 7,\quad r= 464 $$ $$p=3621,\quad q= 5,\quad r= 772 $$ $$p=4779,\quad q= 14,\quad r= 344 $$ $$p=5979,\quad q= 19,\quad r= 316 $$ $$p=14379,\quad q= 16,\quad r= 904 $$ $$p=29559,\quad q= 5,\quad r= 6302 $$ $$p=124419,\quad q= 26,\quad r= 4796 $$ $$p=784419,\quad q= 16,\quad r= 49316 $$ $\endgroup$
    – Alex
    May 2, 2017 at 4:06

1 Answer 1

5
$\begingroup$

If $\dfrac{p^2+7}{q^2-3} = \dfrac{a^2}{b^2}$, where $\gcd(a,b) = 1$, then there is a positive integer $r$ such that $$\begin{cases}p^2 - ra^2 = -7& \\ q^2 - rb^2 = 3. & \end{cases}$$

For each $r$ we have two independent generalized Pell's Equations. There are several algorithms to find the solutions to this equations. I suggest you read the ones here.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .