# 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

• 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 exp onents 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) =
.

 Prev Next

Start solving your Algebra Problems in next 5 minutes!

2Checkout.com is an authorized reseller
of goods provided by Sofmath

Attention: We are currently running a special promotional offer for Algebra-Answer.com visitors -- if you order Algebra Helper by midnight of December 13th you will pay only \$39.99 instead of our regular price of \$74.99 -- this is \$35 in savings ! In order to take advantage of this offer, you need to order by clicking on one of the buttons on the left, not through our regular order page.

If you order now you will also receive 30 minute live session from tutor.com for a 1\$!

You Will Learn Algebra Better - Guaranteed!

Just take a look how incredibly simple Algebra Helper is:

Step 1 : Enter your homework problem in an easy WYSIWYG (What you see is what you get) algebra editor:

Step 2 : Let Algebra Helper solve it:

Step 3 : Ask for an explanation for the steps you don't understand:

Algebra Helper can solve problems in all the following areas:

• simplification of algebraic expressions (operations with polynomials (simplifying, degree, synthetic division...), exponential expressions, fractions and roots (radicals), absolute values)
• factoring and expanding expressions
• finding LCM and GCF
• (simplifying, rationalizing complex denominators...)
• solving linear, quadratic and many other equations and inequalities (including basic logarithmic and exponential equations)
• solving a system of two and three linear equations (including Cramer's rule)
• graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions)
• graphing general functions
• operations with functions (composition, inverse, range, domain...)
• simplifying logarithms
• basic geometry and trigonometry (similarity, calculating trig functions, right triangle...)
• arithmetic and other pre-algebra topics (ratios, proportions, measurements...)

ORDER NOW!