Can you find all natural number solutions of this equation? I tried puting it in wolfram alpha and some other math problem solvers but they just solve it for one solution $$x = 2$$ and $$y = 1$$
$$y^{2} = \frac{24}{49}x + \frac{1}{49}$$
Thanks in advance.
This is just $$49y^2=24x+1$$ $$y^2\equiv1\mod{24}$$ Trying the values of $y$ in the least residue system gives the solutions $$y\equiv\{1,5,7,11,13,17,19,23\}\mod{24}$$ For which the integer value of $x$ is just given by $$x=\frac{49y^2-1}{24}$$ So there are infinitely many natural number solutions of the form $$y=\begin{cases}1+24k\\5+24k\\7+24k\\11+24k\\13+24k\\17+24k\\19+24k\\23+24k\end{cases}$$ $$x=\frac{49y^2-1}{24}$$ Where $k\in\mathbb{Z}$ and $k\ge0$.
$y^2=\frac{24}{49}x+\frac{1}{49}$
Multiply through by $49$: $49y^2=(7y)^2=24x+1$
Rearrange by moving $1$ to LHS: $(7y)^2-1=(7y+1)(7y-1)=24x\Rightarrow x=\frac{(7y+1)(7y-1)}{24}$
In order that $x$ be an integer, $(7y+1)(7y-1)$ must be even (to be divisible by the even number $24$), so $y$ must be odd, and since either $(7y+1)$ or $(7y-1)$ must be divisible by $3$ (since there is a factor of $3$ in $24$), $y$ cannot be divisible by $3$. So pick $y$ from $\{1,5,7,11,13,17,\dots \}$, plug it in to $x=\frac{(7y+1)(7y-1)}{24}$, and integer values for $x$ will pop out.
linear-algebra? – José Carlos Santos Apr 21 '19 at 17:15