# 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 ir reducible 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) = x^{n} +

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) = x^{n} + x^{m} + x^{p}
+ x^{q} + 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 < k_{1} < k_{2} < k_{3}
< n.

• By considering reciprocal polynomials if necessary, we consider m+q ≤n and k_{1}+k_{3}≤n.

• The condition
implies

• Deduce 2n − k_{1} = 2n − q so that k_{1}
= q.

• By adding exponents , deduce 14n + 2k_{3} − 2k_{1} = 14n + 2m −
2q so k_{3} = m.

• Substitute and compare exp onents to obtain

{2n − p, n + p, n + m − p, n + p − q} = {2n − k_{2}, n + k_{2},
n + k_{3} − k_{2}, n + k_{2} − k_{1}}.

• Comparing largest elements of these sets, deduce one of 2n−p and n+p must
equal one of

2n − k_{2} and n + k_{2}.

• If 2n − p = 2n − k_{2} or n + p = n + k_{2}, k_{2} = p
and w(x) = f(x).

• If 2n − p = n + k_{2} or n + p = 2n − k_{2}, then k_{2}
= n − p. Substituting and comparing

exponents, deduce

{n + m − p, n + p − q} = {n + k_{3} − k_{2}, n + k_{2} −
k_{1}} = {m + p, 2n − p − q}.

If n + m − p = m + p, then n = 2p so that k_{2} = n − p = p and w(x) =
f(x). If

n + m − p = 2n − p − q, then n = m + q so that k_{3} = m = n − q, k_{1}
= q = n − m, and

w(x) = .