IMC 2023 · Day 1 · P1

Hint:

• The fixed point of $7x + 1$ is $-1/6$.
• Differentiating twice cancels out the coefficient 49.

Differentiating the equation twice, we get $$\tag{1} f''(7x + 1) = f''(x) \quad \text{or} \quad f''(x) = f''\left( \frac{x - 1}{7} \right).$$ Take an arbitrary $x \in \mathbb{R}$, and construct a sequence by the recurrence $$x_0 = x, \qquad x_{k+1} = \frac{x_k - 1}{7}.$$ By (1), the values of $f''$ at all points of this sequence are equal. The limit of this sequence is $-\frac16$, since $\bigl| x_{k+1} + \frac16 \bigr| = \frac17 \bigl| x_k + \frac16 \bigr|$.

Due to the continuity of $f''$, the values of $f''$ at all points of this sequence are equal to $f''\bigl( -\frac16 \bigr)$, which means that $f''(x)$ is a constant.

Then $f$ is an at most quadratic polynomial, $f(x) = a x^2 + b x + c$. Substituting this expression into the original equation, we get a system of equations, from which we find $a = 36c$, $b = 12c$, and hence $$f(x) = c (6x + 1)^2.$$