How to solve the below system of congruence? I have tried to solve the below congruence system but i don't succeed , such that i have wrote  $3x+5y=6$ as $5y =-3x²\mod (6)$ then i can't  write this  equation as : $y= z \bmod 6$ in order to solve this system , Then my question here is 


Question
    How to solve this system :
    $$
\begin{cases}
3x+5y=6 \\
y \equiv x² \pmod{5}
\end{cases}
$$


 A: We have that


*

*$3x+5y=6 \implies 3x\equiv 1 \mod 5\implies x\equiv 2 \mod 5$

*$y \equiv x² \pmod{5}\implies y \equiv 4 \pmod{5}$


then one solution is given by


*

*$3(2+5k)+5(4+5j)=6\implies 15k+25 j=-20\implies3k+5j=-4\\\implies k=2\quad j=-2$


and therefore


*

*$x=12$

*$y=-6$


Note that since $gdc(3,5)=1$, by Bezout's identity, the equation $3k+5j=-4$ has infinitely many solutions, notably $\forall m\in \mathbb{Z}$
$$3(2)+5(-2)=-4\implies 3(2+5m)+5(-2-3m)=-4 $$
thus all the pairs


*

*$k=2+5m$

*$j=-2-3m$


leads to different solutions


*

*$x=12+25m$

*$y=-6-15m$

A: I would solve first the linear diophantine equation $\;3x+5y=6$. The method is standard:


*

*First step: solve $\;3x+5y=1$.


Start from a Bézout's relation between $3$ and $5$:
$$ 2\times 3-1\cdot 5=1. $$
All other solutions have the form
$$x=2+5k,\quad y=-1-3k\qquad(k\in\mathbf Z).$$


*

*Second step: deduce the solutions of $\;3x+5y=6$.


A basic solution is obtained multiplying the basic solution of the previous equation by $6$: $\; x_0=12,\;y_0=-6$, and the general solution, as in the previous case, is
$$x=12+5k,\quad y=-6-3k\qquad(k\in\mathbf Z).$$


*

*Quadratic congruence: as $x \equiv 2\mod 5$, $\;y\equiv x^2\equiv -1\mod 5$. In terms of $k$, this means
$$y=-6-3k\equiv-1-3k\equiv -1\mod 5 \iff 3k\equiv 0\iff k\equiv 0\mod 5,$$
since $3$ is a unit mod. $5$. Thus $k=5\ell\enspace (\ell\in\mathbf Z)$, and finally
$$x=12+25\ell,\quad y=-6-15\ell\qquad(\ell\in\mathbf Z). $$

A: $3x + 5y = 6$.  As $\gcd(3,5) = 1$, by Bezouts Theorem there are an infinite number of solutions.
$3x = 6 - 5y$
$x = 2 - \frac {5y}3$.  So $3|y$.  Let $y = 3k$ then 
$x = 2 - 5k$.  So all solutions are $\{(2 - 5k, 3k)\}$ 
$y \equiv x^2 \mod 5$
$3k \equiv (2-5k)^2 \mod 5$
$3k \equiv 4 \mod 5$
$2*3k \equiv 2*4 \mod 5$
$k \equiv 3 \mod 5$.
So $k = 2 + 5m$ for some $m$ and $x = 2- 5(2-5m) = -8 + 25m$ and $y = 3(2+5m) = 6 + 15m$.
So all possible solutions are $\{(-8 + 25m, 6+15m)\}$.
