On the diophantine equation xp - x = yq - y

Publicacions Matem`atiques, Vol 43 (1999), 207–216. ON THE DIOPHANTINE EQUATION xp−x=yq−y M. Mignotte†and A. Peth˝ o‡ Abstract We consider the diophantine equation (∗)xp−x=yq−y in integers (x, p, y, q). We prove that for given pand qwith 2 ≤ p<q(∗) has only finitely many solutions. Assuming the abcconjecture we can prove that pand qare bounded. In the special case p= 2 and ya prime power we are able to solve (∗) completely. 1. Introduction. This paper was motivated by the observations of Fiedler and Alford [FA]. We consider the family of diophantine equations (1) xp−x=yq−y, and (except in Section 3, where we consider rational solutions) we consider only non-trivial integral solutions, that is solutions in rational integers (x, y) for which xp−x= 0. Of course we always suppose that |x|, |y|,p,q>1 and that p=q.In[FA], the authors give the following list of positive solutions (x, p, y, q): (3,2,2,3),(6,2,2,5),(15,2,6,3),(16,2,3,5), (13,3,3,7),(91,2,2,13),(280,2,5,7),(4930,2,30,5). †This work was began and finished during two visits of the first author to the University of Debrecen and he wants to thank the people of this University for their kind hospitality. ‡Research supported in part by the Hungarian Foundation for Scientific Research, Grant No. 25157/98. 208 M. Mignotte, A. Peth˝ o Studying systematically (1) we found no other solutions but all the above solutions, as the only solutions of some different families, except the last of the previous list. The exceptional solution (4930,2,30,5) corresponds to the equation x2−x=y5−y. It seems to be a hard problem to solve this equation. We shall first show that non-trivial solutions can exist only when the exponents pand qare coprime. Suppose that pand qare both divisible by some prime ,sayp=pand q=q, then xp−yq=x−y=(xp−yq)  i+j=−1 yiqxjp . If =2,weget|x|p+|y|q≤|x|+|y|, contradiction. If ≥3, then –ifxand yare of the same sign we get again |x|p+|y|q≤|x|+|y|, –ifxy < 0 and y>0 and x<0 then we see that pmust be even and we get for the third time the inequality |x|p+|y|q≤|x|+|y|. The case y<0 and x>0 can be treated similarly. Hence we have proved Proposition 1. If the equation (1) has non-trivial integral solutions then pand qare coprime. Remark. A similar proof shows that equation (1) has no non-trivial solution (x, p, y, q) with y=ztand such that gcd(t, p)>1. We are quite unable to prove a general result about this equation, but we prove some results about the finiteness of the set of solutions when some values among x,y,pand qare fixed. Now we study a certain collections of special cases for which we can obtain some information. Let Sdenote a finite set of primes. The set consisting of 1 and of all those integers which are divisible only by primes belonging to Sis called the set of S-integers. Now we state the finiteness results: 0) If xand pare fixed, then as y|(xp−x), there are only finitely many possible solutions. 1) If x,y∈Zare fixed then xp−yq=x−y. This is a S-unit equation in two unknowns, hence p,q<C(x, y), where the effectively computable function Cdepends only on xand y. (See the book of Shorey-Tijdeman, [Sh-T, Corollary 1.3].) On the diophantine equation xp−x=yq−y209 2) If x,yare S-integers, then (1) becomes a four-term S-unit equation. As none of x,y,xp−x,yq−yand x+yqis zero, this equation has only finitely many solutions (x, p, y, q) by a theorem of Everste [E]. But this result is not effective. 3) If pand yare fixed, then we re-write (1) as Q(x):=xp−x+y=yq. Assume that x0is a multiple root of Q(x). Then x0is also a zero of Q(x)=pxp−1−1, i.e. x0=(1/p)1 p−1ζwhere ζis a (p−1)-th root of unity. As Q(x0) = 0 we obtain y=1 p 1 p−1 ζ−1 pp p−1 ζ=ζ1 p 1 p−11−1 p=ζ1 p 1 p−1 ·p−1 p. The rightmost number cannot be an integer, hence Qis square-free. As p≥2 we can apply Theorem 10.1 of [Sh-T] and conclude that qand x are effectively bounded. In the next two sections we are mainly dealing with the cases pand q fixed. 2. An application of Siegel’s theorem. In this section, we use the following result of Davenport, Lewis and Schinzel [D-L-S]: Theorem A. Let f(x)be a polynomial with integral coefficients of degree n>1and g(y)be a polynomial with integral coefficients of degree m>1.LetD(λ) = discf(x)+λand E(λ) = discg(y)+λ. Suppose that there are at least n/2disctinct roots of D(λ)=0for which E(λ)=0. Then f(x)−g(y)is irreducible over the complex numbers. Further, the genus of the equation f(x)−g(y)=0is strictly positive except possibly when m=2or m=n=3. Apart from these possible exceptions, the equation has at most a finite number of integral solutions. Of course, the last assertion of this theorem is a direct application of the famous result of Siegel [Si] about integral points of curves of positive genus. Results which generalise the previous theorem can be found in more recent papers like [P-S], [R-S] and [Sch]. The book of Stepanov contains also the following more general result: Let n= deg(g), if (n, q) = 1 then the polynomial yq−y−g(x) is absolutely irreducible (see, [St, Corollary, p. 56]). 210 M. Mignotte, A. Peth˝ o For p= 2 equation (1) defines an elliptic or a hyperelliptic curve according to q= 3 and q>3. In both cases the genus of the curve is positive. Hence we may assume 2 <p<qin the sequel. Put f(x)=xp−xand g(y)=yq−y. First we compute the discriminant of the polynomial h(x)=f(x)−λ=xp−x−λ. A common root x of h(x) and h(x) satisfies pxp−x= 0 and xp−x−λ= 0, hence (p−1)x+pλ = 0. This leads to the formula D(λ) := disc(xp−x−λ)= pp(−λ)p−1−(p−1)p−1. Any root λof Dsatisfies |λ|=(p−1)p−p/(p−1). Since the function z→ (z−1)z−z/(z−1) is strictly increasing for z>1 [proof: derivate], the discriminants E(λ) = disc(g(y)−λ) and D(λ) have no common root, and the theorem above applies. We have obtained the following result: Theorem 1. For given pand qwith 2≤p<q, the diophantine equation xp−x=yq−yhas only a finite number of integral solutions. Assuming the abc-conjecture we can prove much more. Let (x, p, y, q) be a non-trivial solution of (1) with min{p, q}=p. Then, as |xp|≈|yq| the abc-conjecture implies |xp|≤(xy(x−y))1+ε<(|x|2+p/q)1+ε, i.e. p= 2 for all but finitely many pairs (p, q). For p= 2 we re-write (1) as (2x−1)2=4yq−(4y−1). Applying again the abc-conjecture we obtain |yq|≈x2≤(x·y2)1+ε, whence q≤4 with finitely many exceptions∗. For the finitely many exceptional pairs (p, q) Theorem 1 implies that max{|x|,|y|} is bounded. Hence we have proved Theorem 2. Assuming the abc-conjecture equation (1) has only finitely many non-trivial solutions. ∗We thank the referee for pointing out to this argument, and to other inaccuraces in an earlier version of this paper. On the diophantine equation xp−x=yq−y211 3. An application of Falting’s theorem. In this section, we look at the set of rational solutions of equation (1). Recall the following proposition (cf. [Fu, Prop. 5, p. 199]): Proposition B. Let Cbe an irreducible plane curve with only ordinary multiple points. Let nbe the degree of C,rP=mP(C). Then the genus of Cis given by g=(n−1)(n−2) 2− P∈C rP(rP−1) 2. We consider the curve Cdefined by equation (1). It follows by Theorem A that this curve is irreducible. In the present case, for the equation xp−x=yq−y, a multiple point (x, y) would be a solution of the system      xp−x=yq−y, pxp−1−1=0, qyq−1−1=0. This implies x=p−1 p−1,y=q−1 q−1 and, after some computation, we get the relation 1 pq−1q(p−1) p(q−1)(p−1)(q−1) =1 qp−1. Now suppose that is a prime number which divides qand, more precisely, suppose that βqand αp−1. Then does not divide pby Proposition 1, and |q−1.Hence (β+α)(p−1)(q−1) = −β(p−1), which is absurd. Thus we have proved that our curve Cdoes not have multiple points. As a consequence, the genus gof Cis equal to g=(q−1)(q−2) 2≥3,when q≥4. [We suppose that q>p, without loss of generality.] Now, by Falting’s theorem [Fa], we know that, there are only a finite nuber of rational points on C. Thus we have proved the following result: 212 M. Mignotte, A. Peth˝ o Proposition 2. The curve Cdefined by the equation xp−x=yq−y has only a finite number of rational points (x, y)when 2≤p<qand q≥4. Remark. In the special case p=2,q= 3, the curve Cis an elliptic curve with rank one, thus it contains an infinite number of rational points. 4. The case p=2. We now consider the special case p= 2. In this special case, multiplying by 4 gives (2) X2=4yq−4y+1, where X=2x−1. Put 4y−1=db2where dis square-free and bis a positive integer. Notice that (2) has the trivial solution (2y−1,y,2). We assume that (2) has the solutions (X,y, q), with qodd and, moreover, that yis a prime power. Put X=2y−1. From the relation X2+db2=4y2, if we put α=1−b√−d 2,β=X+b√−d 2=y−1−b√−d 2, we see that αand βare algebraic integers in the number field K:= Q[√−d], that β=−α2and y=α¯α(where the bar denotes complex conjugation). Clearly gcd(α, ¯α)=1. From the relation X2+db2=4yq,ifγ=X+sb√−d 2, with s=±1, we see that γis an algebraic integer in Kand that the ideal (γ) is equal to (γ)=cq, for some ideal cof the field K. This relation implies (y)=c¯ c. Now we use (for the first time) the fact that y=pffor some prime number p. The decomposition (y)=α¯αproves that psplits in K,say(p)=p¯ pwith pf=(α) and p=¯ p. It is easy to prove that gcd(c,¯ c) = 1. Choosing s suitably we may assume that (γ)=pfq =(α)q. On the diophantine equation xp−x=yq−y213 Now, β=−α2and γ=εαq, where εis a unit in K. Therefore, (3) b√−d=α2−¯α2=εαq−¯ε¯αq. If d= 3 [the case d= 1 cannot occur: d≡−1 (mod 4)], then the only roots of unity in Q(√−d) are ±1, thus we have (4) 1 = U2=|Uq| with Un=(αn−¯αn)/(α−¯α). Consider the special case d= 3. Then the only roots of unity in K are ±1, ±1±√−3 2. Let ε=1+√−3 2. Then αq=γ·1−√−3 2=X+sb√−3 2·1−√−3 2=X+3sb +(sb −X)√−3 4 and so Uq=αq−¯αq α−¯α=(sb −X)√−3 2·1 −b√−3=X−sb 2b. We have Uq∈Zand bodd, hence bdivides X.Thusbdivides also ybecause 4yq=X2+db2, which implies b= 1 and y= 1, which is absurd. Hence, ε=1+√−3 2is not possible. One can exclude similarly the cases ε=1−√−3 2and ε=−1±√−3 2. Hence (2) implies (4) in the case d= 3 too. We consider the equation (4). We have U0= 0 and U1=1. By Theorem 4 of Beukers [B] equation (4) has at most two solutions in q unless y=α¯α= 2, 3 and 5. In our situation we know already two solutions of (4), namely q=1 and 2, hence there are no others in the general case. Consider the exceptional cases following Beukers [B]: If y= 2, then q= 1, 2, 3, 5 and 13, and they give the (already known) solutions (x, p, y, q)=(3,2,2,3),(6,2,2,5),(91,2,2,13). If y= 3, then q= 1, 2 and 5, which corresponds to the (already known) solution (x, p, y, q) = (16,2,3,5). Finally, for y= 5, then q= 1, 2 and 7, and we obtain the (already known) solution (x, p, y, q) = (280,2,5,7). 214 M. Mignotte, A. Peth˝ o Thus, we have proved the following result: Theorem 3. When yis a prime power, the diophantine equation x2−x=yq−y, q > 2, has only the following solutions (x, y, q)=(3,2,3),(6,2,5),(91,2,13),(16,3,5),(280,5,7). We notice that the equation x2=4yq+4y+ 1, where yis a prime power, q≥1, q= 2, has been studied in [T-W1] and [T-W2], in the second paper it is proved that the only solutions are (x, y, q)=(5,3,1) and (11,3,3). In the special case q= 3 we obtain the elliptic diophantine equation x2−x=y3−y. Mordell [M] proved a long time ago that it has the following set of solutions (x, y)=(0,0), (1,0), (0,±1), (1,±1), (3,2), (−2,2), (15,6), (−14,6), which contains four non-trivial solutions. For p= 2, by Proposition 1 the case q= 4 leads only to trivial solutions. Hence the following result: Proposition 3. For p=2and q≤4the only non-trivial integral solutions of equation (1) are (3,2,2,3),(15,2,6,3),(2,2,−2,3),(−14,2,6,3). Remark. We are also able to treat some special cases when p= 3 and yis fixed. Using elliptic curves and the computer algebra system simath one can also prove that for p= 3 and y= 3 the only solution (x, p, y, q) of (1) is (13,3,3,7), which was already found in [FA] and that there are no solutions for p= 3 and y= 2 or 4 ≤y≤8. Here, for each fixed value of y, we have two elliptic equations to consider: Y2=X3−X+ywhen qis even (then Y=yq/2) and yY2=X3−X+y(where Y=y(q−1)/2) when qis odd. 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