For every positive integer n, consider all monic polynomials f(x) with integer coefficients, such that for some real number a

Looks like you're not Brilliant after all.

"lin" also needs to learn how to spell "help" ... and that there is no class called "heeeeeeelp math" -- incredible inability to follow directions.

To find the largest possible number of polynomials f(x) for a fixed n<1000, we need to analyze the given equation:

x(f(x+a)−f(x))=nf(x)

First, let's rewrite the equation by expanding the brackets:

xf(x+a) - xf(x) = nf(x)

Simplifying further:

xf(x+a) - nf(x) = xf(x)

Rearranging the terms:

xf(x+a) - (n+1)f(x) = 0

Now, notice that this is a linear homogeneous differential equation of the form:

y' - (n+1)y = 0

To solve this differential equation, we can express it as a separable equation:

dy/y = (n+1)dx

Integrating both sides:

ln|y| = (n+1)x + C

Applying the exponential function to both sides:

|y| = e^((n+1)x + C)

Since we're looking for monic polynomials, we can discard the absolute value:

y = e^((n+1)x + C)

Now, let's consider the condition that the polynomial has integer coefficients. For this, the term inside the exponential function should be a real number. The expression, (n+1)x + C, should be equal to zero when x is an integer.

(n+1)x + C = 0

Solving for C:

C = - (n+1)x

Now, let's substitute this value of C back into the polynomial equation:

y = e^((n+1)x - (n+1)x)

Simplifying:

y = e^0

y = 1

Hence, the only polynomial that satisfies the given condition for any fixed n<1000 is:

f(x) = 1

Therefore, the largest possible number of such polynomials f(x) is 1.