- MacTutor History of Mathematics

Solution: Day 3, problem 1

The polynomial in question has the form

f(x) = a_{0}x^{n} + ... + a_{n}

where each

a_{i}

is between
0 and 9, the leading coefficient

a_{0}

is positive, and

f(10)

is prime. Suppose that f factors non-trivially as

g(x)h(x)

, where by Gauss's lemma we can suppose that

g(x), h(x)

are in

\mathbb{Z}[x]

.
The primality of

f(10)

implies that one of

g(10)

and

h(10)

, say the former, equals 1 (or - 1, but then replace

g

- g

). Write

g(x) = c (x - b_{1})(x - b_{2}) ... (x - b_{d})

, where

d = \deg(g) ≥ 1

and

c

is an integer.
We claim that each root

b = b_{j}

is in the union of the (closed) left half - plane and the disk of radius 4 centred at the origin. Indeed,

b

is also a root of

f(x)

, so if

Re(b) > 0

and

|b| ≥ 4

then we would obtain a contradiction from

1 ≤ a_{0} ≤ Re(a_{0} + a_{1}b^{-1}) = Re( -a_{2}b^{-2} - ... - a_{n}b^{-n}) ≤ 9|b|^{-2}/(1 - b^{-1}) ≤ {{3}\over{4}}

. It follows that

| 10 - b | ≥ 6

, so

1 = | g(10) | = | c (10 - b_{1}) ... (10 - b_{d}) | ≥ 6^{d}

, which is a contradiction.