All natural number solutions of the equation [closed]

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 $$y = 1$$

$$y^{2} = \frac{24}{49}x + \frac{1}{49}$$

closed as off-topic by José Carlos Santos, John Omielan, Alexander Gruber♦Apr 29 at 1:43

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• You should choose your tags carefully. What has this to do with linear-algebra? – José Carlos Santos Apr 21 at 17:15

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.

• Do you think that all solutions for y are prime numbers? – Adnan C Apr 21 at 18:34
• @Adrian C No. If you continue the list of odd numbers not divisible by $3$, you get numbers including $25,35,49,55$ etc. which are not prime. – Keith Backman Apr 21 at 18:41
• Do you think that $1$ is prime? – Peter Foreman Apr 21 at 19:03
• Not sure who the question is directed to, but see for example this – Keith Backman Apr 21 at 20:39