We show that there exists an effective upper bound for the solutions to the Nagell-Ljunggren equation of the form --=y^{q} in 4 unknowns in integers x>1, y>1, m>2, q>1, when x is a cube of an integer. Our method relies on a refined estimate of linear forms in logarithms

Hirata-Kohno, Noriko

Kovacs-Coskun, Tunde

Miyazaki, Takafumi

Kyoto University Research Information Repository

TitleON THE NAGELL-LJUNGGREN EQUATION (AnalyticNumber Theory : Distribution and Approximation ofArithmetic Objects)Author(s)Hirata-Kohno, Noriko; Kovacs-Coskun, Tunde; Miyazaki,TakafumiCitation数理解析研究所講究録 (2016), 2013: 60-67Issue Date2016-12URL http://hdl.handle.net/2433/231633RightType Departmental Bulletin PaperTextversionpublisherKyoto UniversityON THE NAGELL‐LJUNGGREN EQUATIONN. HIRATA‐KOHNO, T. KOVÁCS AND T. MIYAZAKIABSTRACT. We show that there exists an effective upper bound for the solutions to the Nagell‐x^{m}-1Ljunggren equation of the form --=y^{q} in4 unknowns in integers x>1, y>1,m>2, q>x-11, when x is a cube of an integer. Our method relies on a refined estimate of linear forms inlogarithms.1. INTRODUCTIONIt is a longstanding conjecture that the exponential Diophantine equation in four unknowns,so‐called the Nagell‐Ljunggren equation:(1) \displaystyle \frac{x^{ $\gamma$ n}-1}{x-1}=y^{q} in integers x>1, y>1,m>2, q>1has finitely many solutions (x, y, m, q) . Nagell and Ljunggren confirmed [12][15][16] that apartfrom(2) \displaystyle \frac{3^{5}-1}{3-1}=11^{2}, \frac{7^{4}-1}{7-1}=20^{2}, \frac{18^{3}-1}{18-1}=7^{3},the equation (1) has no solution (x, y, m, q) if either one of the following conditions is satisfied:(i) q=2 , (ii) 3|m , (iii) 4|m , (iv) q=3 and m\not\equiv 5 (mod)6.It remains unknown to date whether the number of the solutions is finite or not, and there isno known solution other than those of (2). It is widely believed that there is no other solution.The problem requires us when it happens a perfect power of an integer to be written with alldigits equal to 1 in base x . Shorey and Tijdeman [22] proved that the equation (1) has onlyfinitely many solutions (x, y, m, q) if one of the following conditions is satisfied:(i) x is fixed, (ii) m has a fixed prime factor, (iii) y has a fixed prime factor. This assertionis effective.It is mentioned by Shorey [20] that the abc conjecture implies the finiteness of the solutionsto the equation (1).Since the case q=2 is solved, there is no loss of generality in assuming that q is an oddprime. The fact that there is no other solution with m even follows from the affirmative answerof Catalans conjecture due to Mihăilescu [14]. Note that it is still an open problem to prove ingeneral the equation (1) has only finitely many solutions of form (x, y, q, q) .2000 Mathematics Subject Classification. 11\mathrm{G}05, 11\mathrm{Y}50.Key words and phrases. Exponential Diophantine equation, the Nagell‐Ljunggren equation.The first author was supported partly by Grant‐in‐Aid for Scientific Research (C), JSPS, no. 26520208 andno. 15\mathrm{K}04799 . The second author was supported partly by the OTKA Grant, 100339 and by the JSPS, P12806.The third author was supported partly by Grant‐in‐Aid for JSPS Research Fellow, JSPS, no. 25‐484.数理解析研究所講究録第2013巻 2016年 60-6760N. HIRATA‐KOHNO, T. KOVÁCS AND T. MIYAZAKINow we consider the Nagell‐Ljunggren equation under the condition that x is a power.Bugeaud, Mignotte, Roy and Shorey [7] proved that the equation (1) has no solution when‐ever x is a square. Hirata‐Kohno and Shorey [9] considered an analogous question when x=z^{ $\mu$}where z>1,  $\mu$\geq 3 and they showed that the equation (1) with x=z^{ $\mu$} with q>2( $\mu$-1)(2 $\mu$-3)has only finitely many solutions effectively bounded depending only on  $\mu$.In this paper, we show that the constant giving an upper bound for the height of the solutions,can be improved using a refinement of a lower bound for the linear forms in logarithms of form|b_{1}\log$\alpha$_{1}+\cdots+b_{n}\log$\alpha$_{n}|.Theorem 1.1 (Hirata‐Kohno, Kovács and Miyazaki). Let z>1 be an integer. Assume  q\neq 5 , 7, 11. Then there exists an effectively computable absolute constant C>0 satisfying thefollowing statement. Suppose (x, y, m, q) is a solution to the equation (1) with x=z^{3} . Then wehave \displaystyle \max(x, y, m, q)\leq C.We may derive the finiteness of the solutions to the equation (1) of Theorem 1.1 from [9],however, our new ingredient here for the proof is based on an advantage of the factor \log E in[11] and [18] appeared in a lower bound for the linear forms in logarithms. We use a lower boundobtained by Bugeaud in [5] which is again precisely calculated by the third author.Note that the result is due to Inkeri when  $\mu$=q=3 (Lemma 4, [10]). Bugeaud and Mignotteproved if  $\mu$=q , there is no solution in (x, y, m, q) (Théorème 9, [6]), and this statement followsfrom a theorem of Bennett on the Thue equation [3] showing, when a>b\geq 1 and n\geq 3 , thatthe equation|ax^{n}-by^{n}|=1has at most 1 solution in positive integers (x, y) (indeed, if  $\mu$=q , we suppose that thereexists z>1, y>1, q\leq 3, m\leq 3 with z^{qm}-1=(z^{q}-1)y^{q} , then consider the equationz^{q}Z^{q}-(z^{q}-1)Y^{q}=1 where Bennetts theorem can bc applied to conclude the statement).In 2007, Bugeaud and Mihăilescu showed  $\omega$(m)\leq 4 if (x, y, m, q) is a solution to the equation(1) [8] and it was improved to  $\omega$(m)\leq 3 by Bennett and Levin [4].2. OUTLINE OF THE PROOFProposition 2.1 (Consequence of Lemma 2 of [9]). The equation (1) with x=z^{3} implies thateither \displaystyle \max(x, y, m, q) is bounded by a positive effective constant, or\displaystyle \frac{z^{7\} $\tau$}-1}{z-1}=y_{1}^{q}, \frac{z^{2m}+z^{ $\tau$ n}+1}{z^{2}+z+1}=y_{2}^{q}where y_{1}>1 and y_{2}>1 are relatively prime integers such that y_{1}y_{2}=y.The next lemma states approximations of certain algebraic numbers by rationals using Padéapproximations found in [5] which is a precise statement of Shorey and Nesterenko [17]. Thisalso improves Lemma 3 of [9].Lemma 2.2. Let A, B, K and n be positive integers such that A>B,K<n, n\geq 3 and $\omega$=(B/A)^{1/n} is not a rational number. For 0< $\phi$<1 , put $\delta$=1+\displaystyle \frac{2- $\phi$}{K}, s=\frac{ $\delta$}{1- $\phi$},u_{1}=40^{n(K+1)(s+1)/(Ks-1)}, u_{2}^{-1}=K2^{K+s+1}40^{n(K+1)}.Assume thatA(A-B)^{- $\delta$}u_{1}^{-1}>1.61ON THE NAGELL‐LJUNGGREN EQUATIONThen| $\omega$-\displaystyle \frac{p}{q}|>\frac{u_{2}}{Aq^{K(s+1)}}for all integers p and q with q>0.Now we apply the lemma above to prove the statement whenever q is fixed. We show:Proposition 2.3. The equation (1) with x=z^{3} and the condition q\neq 5 , 7, 11 implies that max(x, y, m) is bounded by an effectively computable number depending only on q.The proposition 2.3 is proven as follows. Let us consider the equation (1) with x=z^{3}.Recall that Shorey and Tijdeman showed that the equation (1) has only finitely many solutionsif either x is fixed or m has a fixed prime divisor. Then we may assume that \displaystyle \min(m, z)exceeds a sufficiently large constant depending only on q . By Proposition 2.1, we may suppose\displaystyle \frac{z^{m}-1}{z-1}=y_{1}^{q}, \displaystyle \frac{z^{2m}+z^{m}+1}{z^{2}+z+1}=y_{2}^{q} , namely (z-1)y_{1}^{q}=z^{m}-1, (z^{2}+z+1)y_{2}^{q}=z^{2m}+z^{m}+1 . thus0<(z^{2}+z+1)y_{2}^{q}-(z-1)^{2}y_{1}^{2q}\leq 3z^{m} which implies(3) 0<|(\displaystyle \frac{(z-1)^{2}}{z^{2}+z+1})^{1/q}-\frac{y_{2}}{y_{1}^{2}}|<\frac{6z^{m}}{z^{2}y_{1}^{2q}}But Lemma 2.2 with A=z^{2}+z+1, B=(z-1)^{2} gives a contradiction against the upper boundabove.Now it remains to show that the equation (1) with x=z^{3} implies that q is bounded. Theproof uses a lower bound for the linear forms in logarithms.The following result is a precise version of [11, Corollaire 3], whose advantage comes from therole of \log E in a lower bound for the linear forms in logarithms.Proposition 2.4. Let X_{1}/Y_{1} and X_{2}/Y_{2} be multiplicatively independent rational numbers greaterthan the unity. Let b_{1} and b_{2} be positive integers. We consider the linear form=b_{2}\log(X_{2}/Y_{2})-\mathrm{b}_{1}\log(X_{1}/Y_{1}) .Let A_{1}, A_{2} be positive real numbers such that\displaystyle \log A_{i}\geq\max\{\log x_{i}, 1\} (i=1,2) .Let E\geq 3 be a real number such thatE\displaystyle \leq 1+\min\{\frac{\log A_{1}}{\log(X_{1}/Y_{1})}, \frac{\log A_{2}}{\log(X_{2}/Y_{2})}\}.Assume thatE\displaystyle \leq\min\{A_{1}^{3/2}, A_{2}^{3/2}\}.Then we have\log||\geq-35.1(\log A_{1})(\log A_{2})(\log B)^{2}(\log E)^{-3},where\displaystyle \log B=\max\{\log(\frac{b_{1}}{\log A_{2}}+\frac{b_{2}}{1\mathrm{o}gA_{1}})+\log\log E+0.47, 10\log E\}.Using this result, we show the following.62N. HIRATA‐KOHNO, T. KOVÁCS AND T. MIYAZAKICorollary 2.5. Let X_{1}/Y_{1} and X_{2}/Y_{2} be multiplicatively independent rational numbers greaterthan the unity. Assume thatX_{2} 4X_{1}\geq 3, X_{2}\geq 3, \overline{Y_{2}}\leq_{\overline{3}}.Let b_{1} be a positive integer. We consider the linear f_{07 $\gamma \gamma$}b=\log(X_{2}/Y_{2})-b_{1}\log(X_{1}/Y_{1}) .Assume that || is not zero and thatX_{2}/Y_{2}>X_{1}/Y_{1}.Define  $\epsilon$ with  $\epsilon$<1 by\displaystyle \frac{X_{2}}{Y_{2}}=1+\frac{1}{X_{2}^{1- $\epsilon$}}.Then we have\displaystyle \log||\geq-35.1\frac{(\log X_{1})\log X_{2}}{\min\{\log X_{1},(1- $\epsilon$)\log X_{2}\}}(\log b_{1}+10)^{2}.Proof. We may take (A_{1}, A_{2})=(X_{1}, X_{2}) . It suffices to show that we can take E such that(4) \displaystyle \log E=\min\{\log X_{1}, (1- $\epsilon$)\log X_{2}\}.Indeed, if so, since\displaystyle \log(\frac{b_{1}}{\log X_{2}}+\frac{1}{\log X_{1}})+\log\log E\displaystyle \leq\log(\frac{b_{1}}{\log X_{2}}+\frac{1}{\log X_{1}})+\log\log\min\{X_{1}, X_{2}\}\leq\log(b_{1}+1) ,we have(\displaystyle \log B)^{2}(\log E)^{-3}\leq\max\{\log(b_{1}+1)+0.47, 10\log E\}^{2}\cdot(\log E)^{-3}=\displaystyle \max\{\frac{\log(b_{1}+1)+0.47}{\log E}, 10\}^{2}\cdot(\log E)^{-1}\displaystyle \leq\max\{\frac{\log(b_{1}+1)+0.47}{\log 3}, 10\}^{2}\cdot(\log E)^{-1}\leq(\log b_{1}+10)^{2}\cdot(\log E)^{-1}.Hence, Proposition 2.4 gives us the desired inequality. So, we define E by (4). We are left withchecking that all required inequalities on E hold.First, we show E\geq 3 . For this, it suffices to check that X_{2}^{1- $\epsilon$}\geq 3 holds. Since X_{2}/Y_{2}\leq 4/3by our assumption, we haveX_{2}^{ $\epsilon$-1}=X_{2}/Y_{2}-1\leq 1/3.Next, from the definition of  $\epsilon$ , we have\log(X_{2}/Y_{2})=\log(1+X_{2}^{ $\epsilon$-1})<X_{2}^{ $\epsilon$-1},and so-\log\log(X_{2}/Y_{2})>(1- $\epsilon$)\log X_{2}\geq\log E.63ON THE NAGELL‐LJUNGGREN EQUATIONSince X_{1}, X_{2}\geq 3 and X_{2}/Y_{2}>X_{1}/Y_{1} , it follows from (4) that1 + \displaystyle \min\{\frac{\log X_{1}}{\log(X_{1}/Y_{1})}, \displaystyle \frac{\log X_{2}}{\log(X_{2}/Y_{2})}\}>\frac{\log 3}{\log(X_{2}/Y_{2})}>E.Finally, we can observe that (4) yields\displaystyle \log\min\{X_{1}^{3/2}, X_{2}^{3/2}\}=(3/2)\log\min\{X_{1}, X_{2}\}>\log E.This completes the proof. \square We now give an outline of the proof of Theorem 1.1.Proof. Consider the following linear form in two logarithms:=\displaystyle \log(\frac{z^{2}+z+1}{(z-1)^{2}})-q\log(\frac{y_{1}^{2}}{y_{2}}) ,where positive integers y_{1}, y_{2} satisfyy_{1}^{q}=z^{m-1}+z^{m-2}+\displaystyle \cdots+z+1, y_{2}^{q}=\frac{z^{2m}+z^{m}+1}{z^{2}+z+1}.In our case, we may assume thatm is odd >1.Set(X_{1}, Y_{1})=(y_{1}^{2}, y_{2}) , (X_{2}, Y_{2})=(z^{2}+z+1, (z-1)^{2}) , b_{1}=q.Note that X_{1}, X_{2}\geq 3 and\displaystyle \frac{X_{2}}{Y_{2}}=\frac{z^{2}+z+1}{(z-1)^{2}}\leq\frac{4}{3} (\Leftarrow z\geq 11) .(i) Put ($\gamma$_{1}, $\gamma$_{2})=(X_{1}/Y_{1}, X_{2}/Y_{2}) . First, we show that $\gamma$_{2}>$\gamma$_{1} . We already know||=|\displaystyle \log$\gamma$_{2}-q\log$\gamma$_{1}|<\frac{8 $\mu$}{z^{m}}=\frac{24}{z^{m}}.Then\displaystyle \log$\gamma$_{2}>q\log$\gamma$_{1}-\frac{24}{z^{m}}.Hence, since q\geq 3 , it suffices to showq\displaystyle \log$\gamma$_{1}\geq\frac{36}{z^{m}}.64N. HIRATA‐KOHNO, T. KOVÁCS AND T. MIYAZAKIObserve that$\gamma$_{1}^{q}=(\displaystyle \frac{y_{1}^{2}}{y_{2}})^{q}=\frac{y_{1}^{2q}}{y_{2}^{q}}=\frac{(z^{rn-1}+z^{m-2}+\cdots+z+1)^{2}(z^{2}+z+1)}{z^{2m}+z^{rn}+1}>\displaystyle \frac{(z^{m-1}+1)^{2}z^{2}}{z^{2m}+z^{m}+1}=\displaystyle \frac{z^{2m}+2z^{m+1}+z^{2}}{z^{2m}+z^{m}+1}=1+\displaystyle \frac{2z^{m+1}+z^{2}-z^{m}-1}{z^{2m}+z^{m}+1}>1+\displaystyle \frac{1}{z^{m-1}}.(ii)Henceq\displaystyle \log$\gamma$_{1}>\log(1+\frac{1}{z^{rn-1}})>\frac{0.95}{z^{m-1}}>\frac{36}{z^{m}} (\Leftarrow z\geq 38) .Next, we show that X_{1}/Y_{1}, X_{2}/Y_{2} are multiplicative independent. Suppose the contrary.Then, we can find two co‐prime positive integers k, l such that(X_{1}/Y_{1})^{qk}=(Y_{2}/X_{2})^{l},that is,(\displaystyle \frac{(z^{m-1}+z^{m-2}+\cdots+z+1)^{2}(z^{2}+z+1)}{z^{2m}+z^{ $\gamma$ n}+1})^{k}=(\frac{(z-1)^{2}}{z^{2}+z+1})^{l}(5) (z^{2}+z+1)^{k+l}(z^{m-1}+z^{m-2}+\cdots+z+1)^{2k}=(z-1)^{2l}(z^{2rn}+z^{m}+1)^{k}.Since m is odd, we easily see from (5) thatz is even.orSince m\geq 2 and z is even, we have(z^{2}+z+1)^{k+l}\equiv(z+1)^{k+l}\equiv(k+l)z+1 \mathrm{m}\mathrm{o}\mathrm{d} 2z,(z^{rn-1}+z^{m-2}+\cdots+z+1)^{2k}\equiv(z+1)^{2k}\equiv 2kz+1\equiv 1 \mathrm{m}\mathrm{o}\mathrm{d} 2z,(z-1)^{2l}\equiv(z^{2}-2z+1)^{l}\equiv 1 \mathrm{m}\mathrm{o}\mathrm{d} 2z,(z^{2_{7}n}+z^{rn}+1)^{k}\equiv 1 \mathrm{m}\mathrm{o}\mathrm{d} 2z.It follows from (5) thatk+l\equiv 0 \mathrm{m}\mathrm{o}\mathrm{d} 2.This together with the fact \mathrm{g}\mathrm{c}\mathrm{d}(k, l)=1 implies thatk is odd.Now, we reconsider (5). Since k is odd, we may conclude that the termz^{2_{7}n}+z^{m}+1=\displaystyle \frac{z^{3m}-1}{z^{m}-1}has to be a perfect square. But, this contradicts the result of Ljunggren [12].65ON THE NAGELL‐LJUNGGREN EQUATION(iii) We use Corollary 2.5. Noting that $\epsilon$=\displaystyle \frac{\log(\frac{3z(z^{2}+z+1)}{(z-1)^{2}})}{\log(z^{2}+z+1)}(>0.5) ,we have\displaystyle \log||\geq-35.1\max\{\frac{1}{1- $\epsilon$}\log X_{1}, \log X_{2}\}(\log q+10)^{2}=-35.1\displaystyle \max\{\frac{2}{1- $\epsilon$} logy1, \log(z^{2}+z+1)\}(\log q+10)^{2}\displaystyle \geq-35.1\max\{\frac{2}{q(1- $\epsilon$)}\log(\frac{z^{m}-1}{z-1}) , \log(z^{2}+z+1)\}(\log q+10)^{2}>-35.1\displaystyle \max\{\frac{2.1m}{q}\log z, 2.1\log z\}(\log q+10)^{2}=-73.71(\displaystyle \log z)\max\{mq^{-1}, 1\}(\log q+10)^{2}.On the other hand, we know\displaystyle \log||<\log(\frac{24}{z^{m}})=\log 24-m\log z.Combining this with the obtained lower bound for \log|| , we have\displaystyle \log 24-\mathrm{m}\log \mathrm{z}>-73.71(\log z)\max\{mq^{-1}, 1\}(\log q+10)^{2},orm<73.71\displaystyle \max\{mq^{-1}, 1\}(\log q+10)^{2}+\frac{\log 24}{\log z}.If q\leq m , thenq(1-\displaystyle \frac{\log 24}{m\log z})<73.71(\log q+10)^{2},which implies, sayq<40, 000 .If q>m , thenm<73.71(\displaystyle \log q+10)^{2}+\frac{\log 24}{\log z}.Since z^{m}>y_{1^{q}}(=z^{m-1}+z^{m-2}+\cdots+1) , we may replace the left‐hand side above by\displaystyle \frac{\log y_{1}}{\log z}q.Then we have\displaystyle \frac{\log y_{1}}{\log z}q<73.71(\log q+10)^{2}+\frac{\log 24}{\log z}.Hence,q<73.71\displaystyle \frac{\log z}{\log y_{1}}(\log q+10)^{2}+\frac{\log 24}{\log y_{1}}.So, we need an explicit upper estimate of z (or \displaystyle \frac{\log z}{\log y_{1}} ) in terms of q , to bound q . Butthis is already done by Proposition 2.3. This completes the proof of our theorem.\square 66N. HIRATA‐KOHNO, T. KOVÁCS AND T. MIYAZAKIREFERENCES[1] A. Baker, Rational approximations to \sqrt[3]{2} and other algebraic numbers, Quart. J. Math. Oxford, 15, (1964),375‐383.[2] A. Baker, Simultaneous rational approximatzons to certain algebraic numbers, Proc. Cambridge Philos. Soc.,63, (1967), 693‐702.[3] M. A. Bennett, Rational approximation to algebraic number of small height: The diophantine equation|ax^{n}-by^{n}|=1 , J. Reine Angew. Math., 535, (2001), 1‐49.[4] M. A. Bennett and A. Levin, The Nagell‐Ljunggren equation via Runges method, Monatsh. Math., 177,no. 1, (2015), 15‐31.[5] Y. Bugeaud, Linear forms in the logarithms of algebraic numbers close to 1 and applications to Diophantineequations, In : Diophantine Equations, Tata Institute of Fundamental Research, Studies in Mathematics,Narosa Publishing House, 2008, 59‐76.[6] Y. Bugeaud, M. 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Shorey, On the equation z^{q}=\displaystyle \frac{x^{n}-1}{x-1} , Indag. Math., 48, (1986), 345‐351.[20] T. N. Shorey, Exponential Diophantine equations involving products of consecutive integers and relatedequations, In: Number Theory, Hindustan Book Agency, (1999), 463‐495.[21] T. N. Shorey and R. Tijdeman, New Applications of Diophantine approximation to Diophantine equations,Math. Scand., 39, (1976), 5‐18.[22] T. N. Shorey and R. Tijdeman, Exponential Diophantine equations, Cambridge Tracts in Math., 87, 1986,Cambridge Univ. Press.Noriko Hirata‐Kohno Tunde Kovács‐Coskun Takafumi MiyazaklDepartment of Math. Donát Bánki Faculty Faculty of Science & TechnologyCollege of Science & Technology of Mechanical & Safety Engineering Div. of Pure and Applied ScienceNihon University Óbuda University Gunma UniversityTokyo, Chiyoda 1081 Budapest Tenjin‐cho, Kiryu, Gunma101‐8308, Japan Hungary 376‐8518, Japanhirata@math cst nihon‐u.ac jp kovacs.tunde@bgk.uni‐obuda.hu tmiyazaki@gunma‐u.ac jp67

ON THE NAGELL-LJUNGGREN EQUATION (Analytic Number Theory : Distribution and Approximation of Arithmetic Objects)

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ON THE NAGELL-LJUNGGREN EQUATION (Analytic Number Theory : Distribution and Approximation of Arithmetic Objects)

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