No solution to an integer equation

Question

I need to prove that $b^2 - 4ac = 11$ cannot happen for integers $a$, $b$ and $c$. I need to show that $b^2 = 4k + 11$ cannot happen, any hint?

Answer 1

This is easy to prove by a very elementary argument based upon divisibilty; nothing more advanced is required:

With

$b^2 = 4k + 11, \tag 1$

it is clear that $b$ cannot be even, lest $b^2$ is also even; but $4k + 11$ is odd. So, $b$ is odd. Then we have

$\exists n \in \Bbb Z, \;b = 2n + 1; \tag 2$

thus,

$4n^2 + 4n + 1 = b^2 = 4k + 11, \tag 3$

whence

$4n^2 + 4n = 4k + 10 \Longrightarrow 4(n^2 + n - k) = 10 \Longrightarrow 4 \mid 10; \tag 4$

but $4 \mid 10$ is clearly false, whence there is no $b$ such that (1) binds, hence no $b$ such that

$b^2 - 4ac = 11, \tag 5$

no matter how we choose $a, c \in \Bbb Z$.

Answer 2

Just consider the equation modulo $4$. Then $b^2\equiv 3\bmod 4$ is a contradiction, since any square is congruent to $0$ or $1$ modulo $4$.

Reference:

The square of an integer is congruent to 0 or 1 mod 4

Answer 3

I would rewrite $b^2 - 4ac = 11$ as $-4ac = -b^2 + 11$. Then $b$ needs to be odd so that $-b^2 + 11$ can be even. However, with $b$ odd, $b^2 \equiv 1 \pmod 4$, so $-b^2 \equiv 3 \pmod 4$. And since $11 \equiv 3 \pmod 4$ also, $-b^2 + 11 \equiv 2 \pmod 4$, not $0$ like we need it to be for $-4ac$.

Comparing to the other answers, it looks like I took the scenic route to get at the same result.