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.


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    $\begingroup$ You should choose your tags carefully. What has this to do with linear-algebra? $\endgroup$ – 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 $k\ge0$.



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.

  • $\begingroup$ Do you think that all solutions for y are prime numbers? $\endgroup$ – Adnan C Apr 21 at 18:34
  • $\begingroup$ @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. $\endgroup$ – Keith Backman Apr 21 at 18:41
  • $\begingroup$ Do you think that $1$ is prime? $\endgroup$ – Peter Foreman Apr 21 at 19:03
  • $\begingroup$ Not sure who the question is directed to, but see for example this $\endgroup$ – Keith Backman Apr 21 at 20:39

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