So, I need help in showing that for any integers $a$ and $b$, ${{a+b\sqrt-7}\over {2}}$ is the root of a quadratic polynomial with integer coefficients if and only if $a$ is congruent to $b (mod 2)$.
Really stuck on this one, so any help is appreciated.
In general, if $p + qi$ is a root of a polynomial with real coefficients, then so is $p - qi$. So in your case, the two roots are $(a \pm b\sqrt{-7})/2$. That means your equation must be a multiple of $$ \left(x - {{a+b\sqrt-7}\over {2}}\right)\left(x- {{a-b\sqrt-7}\over {2}}\right) $$ which simplifies to $$ x^2 - ax + (a^2 + 7b^2)/4. $$ Multiply through by 4 to get the quadratic $$ 4x^2 - 4ax + (a^2 + 7b^2) = 0. $$
So those two are ALWAYS the roots of a quadratic with integer coefficients. If you add the restriction that the quadratic must be MONIC (i.e., coeff. of $x^2$ is $\pm 1$), then to make the first quadratic be an integer quadratic, you need that $ a^2 + 7 b^2$ is divisible by $4$. It's easy to write $a = 2n+1$ and $b = 2k$ and see that the result is not divisible by 4, or $a = 2n$ and $b = 2k+1$. So for it to be divisible by 4, the two must have the same parity. And if they DO have the same parity, then $a^2 + 7b^2$ IS divisible by 4, so you're done.