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We would like to solve the diophantine

$$7x^2-5y^2=18 \tag{E}$$

We first solve the linear diophantine

$$7x-5y=18$$

Solutions are couples $(7k+2,5k+4)$ where $k$ is an arbitrary integer.

thus $(x,y)$ satisfies (E) iff $x^2 \equiv 2 \pmod 7$ and $y^2\equiv 4 \pmod 5$

Hence $x\equiv 3 , 4 \pmod 7$ and $y\equiv 2 ,3\pmod 5$

i wonder how to continue from here.

thanks for your time.

Bernard
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ahmed
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    I don't think that solving the diophantine equation helps you here. Your iff between between the solutions of (E) and the congruence relations of the squares is actually not an iff but only is an implication from left to right. – Louis Hainaut Mar 20 '20 at 13:57

2 Answers2

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The trick here is to multiply your equation by 7 to obtain $$(7x)^2 - 35y^2 = 126.$$ From here you can first look at the equation $x^2 - 35y^2 = 126$, and then divide $x$ by $7$ to find the solutions of your original equation (note that if $(x, y)$ is a solution, then necessarily $x$ is divisible by $7$). This equation can be solved by the general procedure for Pell equations: you first find the smallest solution, and all other solutions are generated by solutions of $x^2 - 35y^2 = 1$.

Edit: You get the full set of solutions of $x^2 - 35y^2 = 126$ by starting with the solution $(x_0, y_0) = (21, 3)$, and then the smallest solution of $x^2 - 35y^2 = 1$ is $(x, y) = (6, 1)$, so you can build recursively solutions $(x_{n+1}, y_{n+1}) = (6x_n + 35y_n, x_n + 6y_n)$, and in order to find the solutions of your original equation you need to divide $x$ by $7$, which gives you $(x_0, y_0) = (3, 3)$ and $(x_{n+1}, y_{n+1}) = (6x_n + 5y_n, 7x_n + 6y_n)$, and the general theory of Pell equations tells you that these are the only solutions (at least with $x, y$ positive. If you change the sign of $x$ or $y$ it is obviously also a solution).

Louis Hainaut
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$(7x^2-5y^2)=18$, has solution

$(x,y)=((6m-5n),(7m-6n))$

$(m,n)$ is known solution.

$(m,n)=(3,-3)$ we get $(x,y)= (33,39)$

Robert
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  • thanks, are these the only solutions ? – ahmed Mar 21 '20 at 07:03
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    In the formula provided by me, the solutions oscillate between (x,y)=(3,3)=(33,39). Also wolfram alpha site & above user @ Louis Hainaut say's the same thing.But since your equation is pell-like & not a pell equation, developing a formal proof is not going to be easy. – Robert Mar 21 '20 at 16:49