358
The Sum of a Cube and a Fourth Power
Thomas Mautsh and Gerhard J. Woeginger
1
Introduction
The book Problems in Elementary Number Theory (published in Romanian)
by Paul Radovii-Marulesu [2℄ ontains the following nie problem:
Problem A. Prove that the number 1919 annot be written as the sum of a
ube and a fourth power.
A losely related problem was disussed in the summer of 2006 on the
German-language Usenet puzzle newsgroup de.rec.denksport:
Problem B. Do there exist 20 onseutive integers, that an all be written as
the sum of a ube and a fourth power?
Both problems an be settled quite easily by working modulo 13. A
ube leaves one of the residues 0, 1, 5, 8, or 12 modulo 13, while a fourth
power leaves one of the residues 0, 1, 3, or 9 modulo 13. It follows that the
sum of a ube and a fourth power leaves one of the residues 0, 1, . . . , 6; or
8, 9, . . . , 12 modulo 13. Thus, the sum of a ube and a fourth power never
leaves a residue of 7 modulo 13. This settles Problem A, sine the residue of
1919 is 7 modulo 13. It also settles Problem B in the negative, sine at least
one of 20 onseutive integers has a residue of 7 modulo 13.
Working modulo 13 is a good idea in these problems beause modulo 13
there are only a few residue lasses that are ubes and fourth powers. But is
there anything speial about the number 13? Aren't there other moduli that
would work equally well? In this artile, we will answer these questions.
Further, we will show that working modulo any other integer n (whih is not
a multiple of 13) will not solve these problems.
Theorem 1 For eah positive integer n that is not divisible by 13, the ongruene x3 + y 4 ≡ r (mod n) in x and y is solvable for all integers r.
2
Our Tool Kit
The proposition below is a onsequene of the Chinese Remainder Theorem.
Proposition 1 Let r be an integer, and let n1 , n2 , . . . , nℓ be integers that are
relatively prime in pairs. If x3 + y 4 ≡ r (mod n) is solvable in integers for
eah n ∈ {n1 , n2 , . . . , nℓ }, then it is solvable for n = n1 n2 · · · nℓ .
Copyright
c 2008
Canadian Mathematial Soiety
Crux Mathematiorum with Mathematial Mayhem, Volume 34, Issue 6
359
We will also make use of the following ase of Fermat's Little Theorem.
Proposition 2 For all integers x, we have x3 ≡ x (mod 3).
The next proposition is folklore (and a trivial ase of Hensel's Lemma).
It an be derived, for instane, from the results in Chapter 4 of the book by
Ireland and Rosen [1℄.
Proposition 3 Let p be a prime, and let k and r be integers, k 6≡ 0 (mod p)
and r 6≡ 0 (modp), suh that z k ≡ r (mod p) is solvable in integers. Then
z k ≡ r mod pb is solvable in integers for all b ≥ 1.
Our main tool is a speial ase of a famous result of Andre Weil [4℄.
Proposition 4 (A. Weil) Let p be a prime, let k1 and k2 be positive integers,
and let r be an integer with r 6≡ 0 (mod p). Then the number N of solutions
(x, y) of the ongruene xk1 + y k2 ≡ r (mod p) with 0 ≤ x, y ≤ p − 1
satises
|N − p| ≤
3
√
gcd(k1 , p − 1) − 1 · gcd(k2 , p − 1) − 1 · p .
The Proof of the Main Theorem
Now let us turn to the proof of Theorem 1. By Proposition 1, we need only
prove the main theorem for prime powers n = pb , where p is a prime other
than 13. We distinguish four main ases.
Case 1. n = 2b . The pair (x, y) = (1, r − 1) is a solution of the ongruene
x3 + y 4 ≡ r (mod 2). Sine it also satises the ondition
x 6≡ 0 (mod 2),
by Proposition 3 the ongruene x3 ≡ r − y 4 mod 2b with y = r − 1 is
solvable for all integers b ≥ 1.
Case 2. n = 3b . By Proposition 2 the pair (x, y) = (r − 1, 1) is a solution
of the ongruene x3 + y 4 ≡ r (mod 3). Sine it also satises the ondition
y 6≡ 0 (mod 3), by Proposition 3 the ongruene y 4 ≡ r − x3 mod 3b
with x = r − 1 is solvable for all integers b ≥ 1.
Case 3. n = p for some prime p ≥ 5 and p 6= 13. Suppose for the sake of
ontradition that there is an integer r for whih x3 + y 4 ≡ r (mod p) has
no solution. Then r 6≡ 0 (mod p), and Proposition 4 with N = 0 yields the
inequality
√
p ≤
gcd(3, p − 1) − 1 · gcd(4, p − 1) − 1 .
If gcd(3, p − 1) = 1, then we obtain the ontradition p ≤ 0. If we have
gcd(4, p − 1) ≤ 2, then we obtain the ontradition p ≤ 4. The remaining
possibility is that gcd(3, p − 1) = 3 and gcd(4, p − 1) = 4, in whih ase
p ≡ 1 (mod 12). Furthermore the displayed inequality yields p ≤ 36. The
360
only prime with these properties is p = 13, whih yields the nal ontradition. Therefore, x3 + y 4 ≡ r (mod p) is solvable for every integer r.
Case 4. n = pb for some prime p ≥ 5 and p 6= 13. The disussion of Case 3
shows that for every r, there exist x0 and y0 with x30 + y04 ≡ r (mod p).
(a) If x0 6≡ 0 (mod p), then the ongruene x3 ≡ r − y 4 mod pb with
y = y0 is solvable by Proposition 3.
(b) If y0 6≡ 0 (mod p), then the ongruene y 4 ≡ r − x3 mod pb with
x = x0 is solvable by Proposition 3.
() If x0 ≡ 0 (mod p) and y0 ≡ 0 (mod p), then r ≡ 0 (mod p). In
this ase, the ongruene x30 + y04 ≡ r ≡ 0 (mod p) possesses another solution (x0 , y0 ) = (−1, 1), whih redues the proof to the ase
x0 6≡ 0 (mod p) in part (a) above.
This ompletes the analysis of the last ase.
Thus, in all four ases, there is a solution to x3 + y 4 ≡ r mod pb .
The proof of Theorem 1 is omplete.
4
Final Remarks
It is well known that if k1 and k2 are positive integers and p is a prime, then
xk1 + y k2 ≡ r (mod p) is solvable for all integers r exept for nitely many
primes p. Indeed, by Proposition 4 the exeptional primes satisfy
√
p ≤
gcd(k1 , p − 1) − 1 · gcd(k2 , p − 1) − 1 .
From this the reader may verify that eah of the following enumerations of
instanes xk1 + y k2 6≡ r (mod p) is omplete for the respetive exponents
k1 and k2 .
x2
x2
x3
x3
x3
x4
x4
x5
x5
x5
x7
x7
+ y5
+ y9
+ y5
+ y7
+ y8
+ y5
+ y7
+ y7
+ y8
+ y8
+ y8
+ y8
6≡
6
≡
6
≡
6
≡
6
≡
6
≡
6
≡
6
≡
6
≡
6
≡
6
≡
6
≡
7 (mod 11) ;
13, 14 (mod 19) , x2 + y 9 6≡ ±14 (mod 37) ;
±12 (mod 31) ;
±13, ±19 (mod 43) ;
7 (mod 13) ;
7 (mod 11) , x4 + y 5 6≡ ±6, ±12 (mod 41) ;
5, 9, 10, 14, 27 (mod 29) ;
±8, ±10, ±11 (mod 71) ;
7 (mod 11) ,
±6, 8, ±12, 20, 22, 26, 31 (mod 41) ;
5, 9, 10, 14, 27 (mod 29) ,
±21, ±23, ±54 (mod 113) .
361
Ekard Speht [3℄ poses the following nie puzzle in the Problem Corner
of the Mathematial Gazette:
Problem C. Determine all integer solutions (x, y) of x3 + y 7 = 20042008 .
The 2005 USA Mathematial Olympiad onsisted of six hallenging problems. Here is the seond problem from this ompetition:
Problem D. (2005 USAMO) Prove that the system
x6 + x3 + x3 y + y
x3 + x3 y + y 2 + y + z 9
= 147157
= 157147
has no solutions in integers x, y , and z .
And here is a minor variation on the theme of this artile:
Problem E. Prove that there are innitely many integers whih annot be
written in the form x3 − y 4 , where x and y are integers.
We enourage the reader to settle these problems along the lines indiated above. For problem D, it is onvenient to rst add the two equations.
Referenes
[1℄ K. Ireland and M. Rosen, A Classial Introdution to Modern Number
Theory, Springer-Verlag, 1992.
[2℄ P. Radovii-Marulesu, Probleme de Teoria Elementara a Numerelor,
Editura Tehnia, Buharest, 1983.
[3℄ E. Speht, Problem 91.K., Mathematial Gazette 91 (2007) p. 561.
[4℄ A. Weil, Numbers of solutions of equations in nite elds. Bulletin of
the Amerian Mathematial Soiety 55 (1949) pp. 497{508.
Thomas Mautsh
Ettenhauserstrasse 42
CH-8062 Wetzikon ZH
Switzerland
[email protected]
Gerhard J. Woeginger
Dept. of Mathematis and Computer Siene
TU Eindhoven
P.O. Box 513, NL-5600 MB Eindhoven
The Netherlands
[email protected]