Don't want $x$ and $y$ ? Ok, we've $X$ and $Y$ !

$4x^2-22x+15\equiv 0\pmod{13}\implies 4x^2+4x^2+2\equiv 0 \pmod{13}$ $(2x+1)^2\equiv 12 \pmod{13}$ similarly $4y^2+22y+15\equiv 0\pmod{13}\implies (2y-1)^2\equiv 12\pmod{13}$ let the 25 x $A={x_1,x_2,x_3,....,x_{25}}$ .let the 25 y be $B={x_1+1,x_2+1,x_3+1,.....,x_{25}+1}$ . note that if $n^2\equiv 12\pmod{13}, n\equiv 5,8\pmod{13}$ but since n has to be odd, for some integer z : $2x+1=13(2z)+5 , 13(2z+1)+8$ $2x+1=26z+5 ,,26z+21$ $x=13z+2 ,13z+10$ so $X=\sum_{n=1}^{25}(x_n)=\sum_{n=0}^{12} (13n+2)+\sum_{n=0}^{11} (13n+10)$ $X=2018$ and $Y=\sum_{n=1}^{25}(x_n+1)=X+25=2043$ $X+Y=4061$

$4x^2-22x+15 \equiv 0 \pmod{13}\implies 4x^2-9x+2 \equiv 0 \pmod{13}$ Compare the expression $4x^2-9x+2$ with $ax^2+bx+c$ .

The solutions of this will be when we use the substitution

$y=2ax+b$ and $d= b^2-4ac$ ,

and the solutions will be given by $y^2 \equiv d \pmod{13}$

Thus we get $(8x-9)^2 \equiv 9^2-4\times 4\times 2 \equiv 49 \pmod{13}$

This gives $8x-9 \equiv \pm 7 \pmod{13}$ . $8x \equiv 16 \text{ or } 2 \pmod{13}$

This gives $x\equiv 2 \pmod{13}$ and other solution $4x\equiv 1 \pmod{13} \implies x\equiv -3 \equiv 10 \pmod{13}$ .

Thus $x\equiv 2 \text{ or } 10 \pmod{13}$

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Similarly solving for $y$ , we get $y\equiv 3 \text{ or } 11\pmod{13}$

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values of $x$ are of the form $13k +2$ and $13k+10$

values of $y$ are of the form $13k+3$ and $13k+11$

See that they are actually 4 A.P.s (Arithmetic Progressions) with common difference $13$ .

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The sum of first $25$ values of $x$ will be $\displaystyle \color{#3D99F6}{X}= \sum_{k=0}^{12} (2+13k) + \sum_{k=0}^{11} (10+13k)$

The sum of first $25$ values of $y$ will be $\displaystyle \color{#3D99F6}{Y}= \sum_{k=0}^{12} (3+13k) + \sum_{k=0}^{11} (11+13k)$

This gives $\color{#3D99F6}{X+Y}=\boxed{4061}$