How to figure out all even integer values of a quadratic equation?

Question

I know that the answer to the question asked above is infinity, but i have a question like this- (-1)^f(x) = 1, Therefore, i need all even integral values of that f(x). How do I figure this out?

Say f(x)=(x^2)+4x-60

Answer 1

For integer $x$, $$f(x)=(x+2)^2-64$$ is an even number iff $x+2$ is even.

Or

$$x^2+4x-60\equiv x^2\mod2$$ and it is obvious that squaring preserves parity.

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Note that minimum effort would have been to try and observe the easy pattern

$$-60,-57,-48,-39,-28,-15,0\cdots$$

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As noticed by @rishabh, if $x$ is not restricted to be an integer, the possible values are the solutions of the quadratic equations

$$x^2+4x-60=2n.$$

These values are all different, except for $n=0$ (and $x=-10$).

Similarly, the odd values occur at the roots of

$$x^2+4x-60=2n+1.$$

Answer 2

$x^2+4x-60$ can be factorised as $(x-6)(x+10)$.

If $(x-6)(x+10)$ is even, at least one of $x-6$ and $x+10$ must be even. In both cases, $x$ must be even.

Answer 3

You should find the min. value of quadratic $x^{2}+4x-60$ which is $-64$ so $f(x)$ assumes all even numbers in $[-64,\infty)$.If you want to know values of $x$ they are $\frac{-4+\sqrt{16+4×(60+2n)}}{2}$ and $\frac{-4-\sqrt{16+4×(60+2n)}}{2}, n $ an integer in $[-32,\infty)$