ON SOME DIOPHANTINE EQUATIONS (II)

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An. Şt. Univ. Ovidius Constanţa Vol. 10(), 00, 79 86 ON SOME DIOPHANTINE EQUATIONS (II) Diana Savin Abstract In [7] we have studied the equation m n = py, where p is a prime natural number p 3. Using the above result, in this paper, we study the equations c k (x + 6px y +p y ) + pd k (x 3 y + pxy 3 ) = 3z with p {5, 13, 9, 37}, where (c k, d k ) is a solution of the Pell equation c pd = 1. 1. Preliminaries. In order to solve our problems, we need some auxiliary results. Proposition 1.1. ([3], pag.7) The integer solutions of the Diophantine equation x 1 +x +... + x k = x k+1 are the following ones: x 1 = ±(m 1 + m +... + m k 1 m k ) x = m 1 m k...,... x k = m k 1 m k x k+1 = ±(m 1 + m +... + m k 1 + m k ), with m 1,..., m k integer number. From the geometrical point of view, the elements x 1, x,..., x k are the sizes of an orthogonal hyper-parallelipiped in the space R k and x k+1 is the length of its diagonal. Proposition 1.. ([1], pag.150) For the quadratic field Q( d), where d N, d is square free, its ring of integers A is Euclidian with respect to the norm N, in the cases d {, 3, 5, 6, 7, 11, 13, 17, 19, 1, 9, 33, 37, 1, 57, 73}. Proposition 1.3.([1], pag11) Let K=Q( d) be a quadratic field with A as its ring of integers. For a A, a U(A) if and only if N(a)=1. Key Words: Diophantine equation; Pell equation. 79

80 Diana Savin Proposition 1.. ([7],Theorem 3..). Let p be a natural prime number greater than 3.If the equation m n = py has a solution m, n, y Z, then it has an infinity of integer solutions. Results. Proposition.1. The equation m n = 5y solutions. has an infinity of integer Proof. The equation m n = 5y has nontrivial integer solutions, for example m = 5, n = 155, y = 600. Following Proposition 1.., the equation m n = 5y has an infinity of integer solutions. Proposition.. The equation m n = 13y solutions. has an infinity of integer Proof. It is sufficient to show that the equation m n = 13y has nontrivial integer solutions. In deed m = 1779, n = 8099, y = 157960 is such a solution.by Proposition 1.., the equation m n = 13y has an infinity of integer solutions. Now, we study our equations for p {5, 13}. Proposition.3. The equations c k (x + 6px y +p y ) + pd k (x 3 y + pxy 3 ) = 3z, with p {5, 13}, where (c k, d k ) is a solution of the Pell equation c pd = =1, have an infinity of integer solutions. Proof. If p {5, 13}, then p 5 (mod 8). By Proposition 1.., the ring A of the integers of the quadratic field Q( p) is Euclidian with respect ] to the norm N. But p 5 (mod 8) implies p 1 (mod )and A=Z. [ 1+ p We shall study the equation m n = py, where p is a prime number, p 5 (mod 8) and ( m, n ) = 1, in the ring A. The equation m n = py is equivalent with (m y p)(m + y p) = n. Let α A be a common divisor of m py and m + py. As α A, α = c + d p, c, d Z, and c, d are simultaneously even or odd. As α/(m + y p) and α/(m y p), we have α/m and α/y p, therefore N(α)/m (in Z) and N(α)/py (in Z), hence N(α)/(m, py ). (m, n) = 1 implies (m, y) = 1 (if (m, y) = d > 1 then m

On Some Diophantine Equations (II) 81 and n would not be relatively prime). Analogously, (m, p) = 1 implies in turn that (m, py ) =, hence N(α) {1,, }. If N(α) =, then c p d =. If c d p =, then c pd = 8, c, d Z and c, d are simultaneously even or odd. If c and d are odd numbers, then c, d 1 ( mod 8). But p 5 ( mod 8). Then c pd ( mod 8), which implies that the equation c pd = 8 does not have integer solutions. If c and d are even numbers, then let us take them c = c, d = d,with c, d Z. We get c pd = 8, then (c ) p(d ) =. But p 5 ( mod 8) implies: (c ) p(d ) ( mod 8), if c, d are odd numbers, (c ) p(d ) 0 or ( mod 8), if c, d are even numbers, (c ) p(d ) = an odd number, if c, d are one even and the other odd. Therefore the equation (c ) p(d ) = does not have integer solutions. If c d p =, that means c pd = 8, with c, d Z + 1 or c, d Z. If c and d are odd numbers, then c, d 1 ( mod 8). As p 5 ( mod 8), this implies c pd ( mod 8), which gives us that the equation c pd = 8 does not have integer solutions. If c and d are even numbers, then c = c, d = d, c, d Z. We get c pd = 8,which means that (c ) p(d ) =. But, as above, p 5 ( mod 8) implies that the equation (c ) p(d ) = does not have integer solutions.we get N(α). If N(α) =, then c p d c =. If p d =, then c pd = 16, where c, d Z and c, d are simultaneously either even or odd. If c and d are odd numbers, then c, d 1 ( mod 8), and, since p 5 ( mod 8), c pd ( mod 8),which implies that the equation c pd = 16 does not have integer solutions. If c and d are even numbers, then c = c, d = d, with c, d Z, therefore (c ) p(d ) =. This equation may have integer solutions only if c, d are simultaneously either even or odd. The equation (c ) p(d = ( ) ( ) ( c is equivalent with d = 1. If we denote α c = + ) d p A, with c, d Z + 1 or c, d Z, we get α U (A).From α = c + d p, we obtain that α = α, α U (A). Supposing that is reducible in A,hence there exist a1 + b1 p, a + b p A ( a1, a, b 1, b Z, a 1, b 1, as well as, a, b being simultaneounsly odd or even) such that = ( a1 +. b1 p)( a + b p),hence N() = N( a 1 + b1 p)n( a + b p). This is equivalent with = N( a1 + b1 p)n( a + b p). But we have previously proved that there aren t elements in A having the norm equal with. We get N( a1 + b1 p) = 1

8 Diana Savin or N( a + b p) = 1, therefore a 1 + b1 p U (A)or a + b p U (A), hence is irreducible in A. We come back to the fact that α/(m + y p and α/(m y p. This implis α /(m + y p) and α /(m y p), then /(m + y p) and /(m y p),therefore /(m py ). This means /n. As is irreducible in A, we get /n ( in A), hence /n. This is equivalent with /(m + y p) (m y p) ( in A), which implies k /(m + y p)or k /(m y p), k N,k. As k /(m + y p), k N,k, implies /(m + y p), hence there exists ( a + ) b p A ( either a, b Z + 1 or a, b Z ) such that m + y p = ( a + ) b p, hence m = a and y = b (in Z), then /m and /y (in Z). As m n = py, this implies /n (in Z), in contradiction with the fact that (m, n) = 1. Analogously we get to contradiction in the case of the equation c p d =. Therefore N(α). From the previously proved, N(α) and N(α), hence N(α) = 1 and α U (A). We obtained that (m + y p) and (m y p) are relatively prime elements in A, but (m y p)(m + y p) = n, therefore there exists ( f + g ) p A with the property: m +y p = ( c k + d ( k p) f + g ) p, ( c k + d k p) U (A) ( here ck, d k Z, c k, d k are simultaneously odd or even, N( c k + d k p) = 1).This is equivalent to m + py = = ( c k + d ) ( k f 16 + f 3 g p + 3f g p 8 + fg3 p ) p + g p 16, which is equivalent to 3(m + y p) = (c k + d k p)(f + f 3 g p + 6f g p + fg 3 p p + g p ), implying the system: { 3m = c k f + 6pc k f g + p c k g + pf 3 gd k + p fg 3 d k 3y = c k f 3 g + pc k fg 3 + d k f + 6pd k f g + p d k g, equivalently { 3 m = c k (f + 6pf g + p g ) + pd k (f 3 g + pfg 3 ) 3y = d k (f + 6pf g + p g ) + c k (f 3 g + pfg 3 ). We have already proved that the equation m n = py, where p {5, 13} has an infinity of integer solutions. Therefore, if the system: { 3 m = c k (f + 6pf g + p g ) + pd k (f 3 g + pfg 3 ) 3y = d k (f + 6pf g + p g ) + c k (f 3 g + pfg 3 ) has an infinity of integer solutions and hence the equation 3 m = c k (f + 6pf g + p g ) + pd k (f 3 g + pfg 3 ) has an infinity of integer solutions.

On Some Diophantine Equations (II) 83 We do not know if the Diophantine equation m n = py has nontrivial solutions for p {9, 37}, but we may prove the following result. Proposition.. If the equations c k (x + 6px y +p y ) + pd k (x 3 y+ +pxy 3 ) = 3z,with p {9, 37}, where (c k, d k ) is a solution of the Pell equation c pd = 1,have a solution x, y, z Z, then they have an infinity of integer solutions. Proof. In our case again p 5 (mod 8) and the ring A of the integers of the quadratic field Q( p) is Euclidian with respect to the norm N, A ] being Z. We study the equation m n = py, where p is prime [ 1+ p number, p 5 (mod 8) in the ring A.The given equation is equivalent to (m y p)(m + y p) = n. Let α A be a common divisor of m y p and m + y p. Then α = c + d p, with c, d Z + 1 or c, d Z. As α/(m + y p) and α/(m y p), we have α/m and α/y p, so N(α)/m and N(α)/py ( in Z ), hence N(α)/(m, py ).(m, n) = 1 implies (m, y) = 1 (if (m, y) = d > 1, then m and n are not relatively prime). Analogously, (m, p) = 1 implies in turn that (m, py ) = 1, (m, py ) =, hence N(α) {1,, }. If N(α) =, we have c p d =. If c d p =, that means c pd = 8, c, d Z and c, d are simultaneously even or odd. If c and d are odd numbers, then c, d 1 ( mod 8). As p 5 ( mod 8),then c pd ( mod 8 ), which implies that the equation c pd = 8 does not have integer solutions. If c and d are even numbers then c = c, d = d, c, d Z. We get c pd = 8, therefore (c ) p(d ) =. But p 5 ( mod 8) implies:(c ) p(d ) ( mod 8), if c, d are odd numbers,(c ) p(d ) 0 or ( mod 8), if c, d are even numbers, (c ) p(d ) = an odd number, if c, d are one even and another odd, and the equation (c ) p(d ) = does not have integer solutions. If c p d =, then c pd = 8, c, d Z or c, d Z + 1. If c and d are odd numbers, then c, d 1 ( mod 8). But p 5 ( mod 8) impliesc pd ( mod 8), which gives us that the equation c pd = 8 does not haveinteger solutions. If c and d are even numbers, then c = c, d = d, c, d Z. We get the equation (c ) p(d ) =, which does not have integer solutions. We get N(α). In the same way as

8 Diana Savin above, we prove that N(α). It remains only α U (A) and m + y p, m y p are relatively prime elements in A. As (m y p)(m + y ( p) = n, there exists f + g ) p A such that: m +y p = ( c k + d k p) ( f + g p ), ( c k + d k p) U (A) (ck, d k Z, c k, d k are simultaneously even or odd, N( c k + d k p) = 1), which is equivalent tom + y ( p = ( c k + d k ) f 16 + f 3 g p + 3f g p 8 + fg3 p ) p + g p 16,which is equivalent to 3(m + y p) = (c k + d k p)(f + f 3 g p + 6f g p + fg 3 p p + g p ). This implies the system: { 3m = c k f + 6pc k f g + p c k g + pf 3 gd k + p fg 3 d k 3y = c k f 3 g + pc k fg 3 + d k f + 6pd k f g + p d k g, which implies the system: { 3 m = c k (f + 6pf g + p g ) + pd k (f 3 g + pfg 3 ) 3y = d k (f + 6pf g + p g ) + c k (f 3 g + pfg 3 ). We have already proved that, if the equation m n = py has a nontrivial solution inz, it has an infinity of integer solutions. Therefore, if the system { 3 m = c k (f + 6pf g + p g ) + pd k (f 3 g + pfg 3 ) 3y = d k (f + 6pf g + p g ) + c k (f 3 g + pfg 3 ), has a nontrivial solution in Z, it has an infinity of integer solutions. Therefore, if the equation 3 m = c k (f + 6pf g + p g ) + pd k (f 3 g + pfg 3 ) has a nontrivial solution in Z, it has an infinity of integer solutions. References [1] T. Albu, I.D.Ion, Chapters of the algebraic theory of numbers ( in Romanian), Ed. Academiei, Bucureşti, 198. [] V. Alexandru, N.M. Goşoniu, The elements of the theory of numbers ( in Romanian), Ed.Universităţii Bucureşti, 1999. [3] T. Andreescu, D. Andrica, An Introduction in the Study of Diophantine Equations (in Romanian), Ed. Gil, Zalău, 00. [] L.J. Mordell, Diophantine Equations, Academic Press, New York,1969.

On Some Diophantine Equations (II) 85 [5] L. Panaitopol, A. Gica, An introduction in arithmetic and the theory of numbers( in Romanian),Ed. Universităţii Bucureşti, 001. [6] Ireland, M. Rosen, A Classical Introduction to Modern Number Theory,Springer- Verlag, 199. [7] D. Savin, On Some Diophantine Equations ( I )Analele Universităţii Övidius,Constanţa, ser. Matematica, 11 (00), fasc.1, pag. 11-13. [8] W.Sierpinski, What we know and what we do not know about prime numbers( in Romanian), ( transl. from the Polish edition,196), Bucureşti. [9] C. Vraciu, M. Vraciu, The elements of Arithmetic, Ed. All, Bucureşti, 1998. Ovidius University of Constanta Department of Mathematics and Informatics, 90057 Constanta, Bd. Mamaia, 1 Romania e-mail: Savin.Diana@univ-ovidius.ro

86 Diana Savin

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