Are you smarter than an 8th grader? #2

You should really edit the image. There are people who know Japanese in Brilliant, including me :)

Anyway, here's the solution:

If $x$ is an integer then suppose that $x=14q_1+r_1$ for $q_1$ and $r_1$ being integers and $0\le r_1<14$ .

Also suppose that $a=14q_2+6$ and $b=14q_3+1$ for $q_2$ and $q_3$ being integers.

We have that:

$x^{ 2 }-2ax+b=0$ $\Leftrightarrow (14q_{ 1 }+r_{ 1 })^{ 2 }-2(14q_{ 2 }+6)x+14q_{ 3 }+1=0$ $\Leftrightarrow 14(14q_{ 1 }^{ 2 }+2q_{ 1 }r_{ 1 })+r_{ 1 }^{ 2 }+14(-2x)-12x+14q_{ 3 }+1=0$ $\Leftrightarrow 14(14q_{ 1 }^{ 2 }+2q_{ 1 }r_{ 1 }-2x+q_{ 3 })+r_{ 1 }^{ 2 }-12(14q_{ 1 }+r_{ 1 })+1=0$ $\Leftrightarrow 14(14q_{ 1 }^{ 2 }+2q_{ 1 }r_{ 1 }-2x+q_{ 3 })+14(-12q_{ 1 })-14r_{ 1 }+2r_{ 1 }+1=0$ $\Leftrightarrow 14(14q_{ 1 }^{ 2 }+2q_{ 1 }r_{ 1 }-2x+q_{ 3 }-12q_{ 1 }-r_{ 1 })+(r_{ 1 }+1)^{ 2 }=0$

Obviously $14(14q_{ 1 }^{ 2 }+2q_{ 1 }r_{ 1 }-2x+q_{ 3 }-12q_{ 1 }-r_{ 1 })\equiv 0\quad (mod\quad 14)$ and $0 \equiv 0 \quad (mod \quad 14)$ .

Hence $(r_{ 1 }+1)^{ 2 }\equiv 0\quad (mod\quad 14)$ .

This leads to $(r_{ 1 }+1)^{ 2 }\equiv 0\quad (mod\quad 2)$ and $(r_{ 1 }+1)^{ 2 }\equiv 0\quad (mod\quad 7)$ .

Both lead to $r_{ 1 }+1\equiv 0\quad (mod\quad 2)$ and $r_{ 1 }+1\equiv 0\quad (mod\quad 7)$ (since $2$ and $7$ are primes), or $r_{ 1 }+1\equiv 0\quad (mod\quad 14)$ . And since $0\le r_1<14$ , $r_1=13$ .

After all of the messy work we've done, we can proudly announce that the remainder of $x$ when divided by $14$ is $\boxed{13}$ .

Look at the equation mod 14,

$x^2 - 2(6)x + (1) = x^2 + 2x + 1 = (x+1)^2 = 0.$

Therefore x is -1 = 13 (mod 14).

Solving the equation $x^2-2ax+b=0$ , we get $x=\frac{2a\pm\sqrt{4a^2-4b}}{2}=a\pm\sqrt{a^2-b}$ In order for $x$ to be an integer, $a^2-b=k^2$ . This implies that: $k^2\equiv7\pmod{14}\Rightarrow k\equiv7\pmod{14} \Rightarrow x\equiv6\pm7\equiv13\pmod{14}$