Double tangent of a quartic

Start with the tangency condition: $f'(k) = f'(-k) = 1.$ So $f'(x)-1$ is divisible by $x-k$ and $x+k.$ Since $f'$ is a cubic, this implies that $f'(x) = (x^2-k^2)(4ax+3b)+1,$ for some $a,b.$ (I put the $4$ and the $3$ in front to make life easier later.)

Taking the antiderivative, we get $f(x) = ax^4+bx^3-2ak^2x^2+(1-3bk^2)x+C$ for some constant $C.$

Now $f(k) + f(-k) = 4,$ which gives $2ak^4-4ak^4+2C = 4,$ so $C = ak^4+2.$ Plugging back in, we get $f(x) = ax^4+bx^3-2ak^2x^2+(1-3bk^2)x+(ak^4+2) = a(x^4-2k^2x^2+k^4) + bx(x^2-3k^2) + x + 2.$

Now $f(k) = k+2,$ so plugging in gives $k+2 = -2bk^3+k+2,$ so since $k \ne 0,$ $b$ must be $0.$ Thus $f(x) = a(x^4-2k^2x^2+k^4)+x+2,$ so $f(0) = ak^4+2.$ Note that $a$ is the positive integer leading coefficient and $k$ is a positive integer $>1.$

When $k=2,$ there are $126$ integers of the form $16a+2$ in the range $[0,2018].$

When $k=3,$ there are $24$ integers of the form $81a+2$ in that range. Note that one of them is also of the form $16a+2,$ namely $81\cdot 16 + 2 =1298.$

When $k=4,$ the integers of the form $256a+2$ are already in the list of integers of the form $16a+2,$ so there are no new ones.

When $k=5,$ there are $3$ integers of the form $625a+2.$ None of them overlap with the previous lists.

When $k=6,$ there is one positive integer of the form $1296a+2$ in the range, namely $1298,$ but it was already in the list of integers of the form $16a+2.$

When $k \ge 7,$ $ak^4+2$ is too large to be in the range.

So there are a total of $126 + 23 + 3 = \fbox{152}$ possible values of $f(0)$ in the range $[0,2018].$

This problem is trivial. Suppose p(x) is our quartic function. Then we can substitute x = abs(k) into p(x) -> p(abs(k)), solving the equations is simple casework. :P

P.S. Orgo 2 was also trivial!