Ljunggren's Approach to Specific Lacunary Results
Theorem (Ljunggren): Let n and m be integers with n
> m > 0, and let ∈ {1,−1} for
∈ {1, 2}. Then the non-cyclotomic part of
is irreducible or identically 1.
Proof:
• The non- reciprocal part of f(x) = is the
same as the non-cyclotomic part of
f(x) (consider
• Suppose w(x) ∈ with
The goal is to show w(x) = ±f(x) or
w(x) = ±. This will
imply the non- reciprocal ( equivalently , non-cyclotomic) part of
f(x) is irreducible or 1.
• We can suppose w(x) has positive leading coefficient and m ≤n − m (the latter
by using
instead of f if necessary).
• Observe that w(0) ≠ 0 and w(x), , f(x),
and have the same degree, namely n.
• Since each coefficient of w (x) is either 1
or −1. Write w(x) = xn +
where
• We can suppose k≤ n − k.
• Note that
and
• Comparing the least two exponents above,
,and k = m. Thus, w(x) = f(x).
Theorem (F. & Solan): Let f(x) = xn + xm + xp
+ xq + 1 be a polynomial with n > m > p >
q > 0. Then the non- reciprocal part of f(x) is either irreducible or 1.
Proof:
• Suppose w(x) ∈ with
The goal is to show w(x) = ±f(x) or
w(x) = ±
• In this case, we may further suppose w(x) is a 0, 1- polynomial (and do so).
Write w(x) =
with 0 < k1 < k2 < k3
< n.
• By considering reciprocal polynomials if necessary , we consider m+q ≤n and k1+k3≤n.
• The condition
implies
• Deduce 2n − k1 = 2n − q so that k1
= q.
• By adding exponents , deduce 14n + 2k3 − 2k1 = 14n + 2m −
2q so k3 = m.
• Substitute and compare exponents to obtain
{2n − p, n + p, n + m − p, n + p − q} = {2n − k2, n + k2,
n + k3 − k2, n + k2 − k1}.
• Comparing largest elements of these sets, deduce one of 2n−p and n+p must
equal one of
2n − k2 and n + k2.
• If 2n − p = 2n − k2 or n + p = n + k2, k2 = p
and w(x) = f(x).
• If 2n − p = n + k2 or n + p = 2n − k2, then k2
= n − p. Substituting and comparing
exponents , deduce
{n + m − p, n + p − q} = {n + k3 − k2, n + k2 −
k1} = {m + p, 2n − p − q}.
If n + m − p = m + p, then n = 2p so that k2 = n − p = p and w(x) =
f(x). If
n + m − p = 2n − p − q, then n = m + q so that k3 = m = n − q, k1
= q = n − m, and
w(x) = .