<s>[INST] If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11) If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11)
Not sure if I'm approaching this problem correctly, any help is appreciated.
So far I have:
$7x^2 - 4x + 1 \equiv 0 $ (mod 11)
$21x^2 - 12x + 3 \equiv 0 $ (mod 11)
$-x^2 - x + 3 \equiv 0 $ (mod 11)
$-x^2 - x  \equiv -3 $ (mod 11)
$-1(x^2 + x)  \equiv -3 $ (mod 11)
$(x^2 + x)  \equiv 3 $ (mod 11)
-Not sure if I am heading the right way, once again any help is appreciated.
 [/INST] Multiply by $8$ (which is the inverse of $7$ mod $11$) and get: $x^2+x+8=0$ (mod $11$).
Then complete the square. "Half of 1" is the same as "half of 12" (mod 11) so we can rewrite our equation as $(x+6)^2-36+8=0$ (mod $11$) which simplifies to $(x+6)^2=6$ (mod $11$).
So your equation has a solution if and only if 6 is a square (mod 11). 
It isn't, so there is no solution. 
To show that 6 is not a square (directly) consider: $0^2=0$, $(\pm 1)^2=1$, $(\pm 2)^2=4$, $(\pm 3)^2=9$, $(\pm 4)^2=16=5$, and $(\pm 5)^2 = 25=3$ (all mod $11$).
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