Study on certain results of exponential diophantine equations

This paper reveals that let p be a fixed odd prime. Using certain results of exponential Diophantine equations, we prove that (i) if p ≡ ±3(mod  8), then the equation 8x + py = z2 has no positive integer solutions (x, y, z); (ii) if p ≡ 7(mod  8), then the equation has only the solutions (p, x, y, z) = (2q − 1, (1/3)(q + 2), 2, 2q + 1), where q is an odd prime with q ≡ 1(mod  3); (iii) if p ≡ 1(mod  8) and p ≠ 17, then the equation has at most two positive integer solutions (x, y, z).

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Introduction

Let Z, N be the sets of all integers and positive integers, respectively. Let p be a fixed odd prime. Recently, the solutions (x, y, z) of the equation 8x+py=z2, x,y,z∈N (1) were determined in the following cases: 1. (Sroysang [1]) if p = 19, then (1) has no solutions; 2. (Sroysang [2]) if p = 13, then (1) has no solutions; 3. (Rabago [3]) if p = 17, then (1) has only the solutions (x, y, z) = (1,1, 5), (2,1, 9), and (3,1, 23). In this paper, using certain results of exponential Diophantine equations, we prove a general

Result

as follows. Theorem 1. If p ≡ ±3(mod 8), then (1) has no solutions (x, y, z). If p ≡ 7(mod 8), then (1) has only the solutions (p,x,y,z)=(2q−1,q+23,2,2q+1), (2) where q is an odd prime with q ≡ 1(mod 3).

If p ≡ 1(mod 8) and p ≠ 17, then (1) has at most two solutions (x, y, z). Obviously, the above theorem contains the results of [1, 2]. Finally, we propose the following conjecture. Conjecture 2.

If p ≠ 17, then (1) has at most one solution (x, y, z). Preliminaries Lemma 3. If 2n − 1 is a prime, where n is a positive integer, then n must be a prime. Proof See Theorem 1.10.1 of [4].

Lemma 4. If p is an odd prime with p ≡ 1(mod 4), then the equation u2−pv2=−1, u,v∈N (3) has solutions (u, v). Maths 2017; 2(5): 22-24 Dr. Sajjad Zahir Associate Professor, Department of Statistics, D.N.

College, Masaurhi, Bihar, India Correspondence Dr. Sajjad Zahir Associate Professor, Department of Statistics, D.N. College, Masaurhi, Bihar, India Proof Using the same method as in the proof of Lemma7, we can obtain this lemma without any difficulty. Lemma 9.

If D belongs to the exceptional case (ii), then (6) has at most one solution (X, n) with 3∣n. Proof Notice that, for any positive integer r, there exists at most one number of 5, 2r + 3, and 4r + 7 which is a multiple of 3. Thus, by Lemma6, the lemma is proved. Lemma 10.

The equation Xm−Yn=1, X,Y,m,n∈N, min{X,Y,m,n}>1 (11) has only the solution (X, Y, m, n) = (3,2, 2,3). Proof See [8]. Proof of Theorem We now assume that (x, y, z) is a solution of (1). Then we have gcd(2p, z) = 1.

If 2∣y, since gcd(z + py/2, z − py/2) = 2, then from (1) we get z+py/2=23x−1, z−py/2=2, (12) where we obtain z=23x−2+1, (13) py/2=23x−2−1. (14) Since p > 1, applying Lemma 10 to (14), we get y=2, p=23x−2−1. (15) Further, by Lemma 3, we see from the second equality of (15) that p=2q−1, q=3x−2 (16) is an odd prime with q ≡ 1(mod 3). Therefore, by (13), (15), and (16), we obtain the solutions given in (2). Obviously, if p satisfies (2), then p ≡ 7(mod 8). Otherwise, since 2∤y, we see from (1) that p ≡ py ≡ z2 − 8x ≡ 1(mod 8).

It implies that if p ≡ ±3(mod 8), then (1) has no solutions (x, y, z). If p ≡ 7(mod 8), then (1) has only the solutions (2). Here and below, we consider the remaining cases that p ≡ 1(mod 8). By the above analysis, we have 2∤y.

If y > 1, then y ≥ 3 and (4) has the solution (X, Y, m, n) = (z, p, 3x, y) with 3∣m. But, by Lemma 5, it is impossible. Therefore, we have y=1. (17) Substituting (17) into (1), the equation X2−p=2n, X,n∈N, n>2 (18) has the solution (X, n) = (z, 3x) with 3∣n. Since p ≡ 1(mod 8), by Lemma 4, (3) has solutions (u, v).

Therefore, Proof See Section 8.1 of [5]. Lemma 5. The equation X2−2m=Yn, X,Y,m,n∈N, gcd(X,Y)=1, Y>1, m>1, n>2 (4) has only the solution (X, Y, m, n) = (71,17,7, 3). Proof See Theorem 8.4 of [6].

Lemma 6. Let D be a fixed odd positive integer. If the equation u2−Dv2=−1, u,v∈N (5) has solutions (u, v), then the equation X2−D=2n, X,n∈N, n>2 (6) has at most two solutions (X, n), except the following cases: 1. D = 22r − 3 · 2r+1 + 1, (X, n) = (2r − 3,3), (2r − 1, r + 2), (2r + 1, r + 3), and (3 · 2r − 1,2r + 3), where r is a positive integer with r ≥ 3; 2.

D = ((1/3)(22r+1 − 17))2 − 32, (X, n) = ((1/3)(22r+1 − 17), 5), (1/3)(22r+1 + 1,2r + 3), and ((1/3)(17 · 22r+1 − 1), 4r + 7), where r is a positive integer with r ≥ 3; 3. D = 22r1 + 22r2 − 2r1+r2+1 − 2r1+1 − 2r2+1 + 1, (X, n) = (2r2 − 2r1 − 1, r1 + 2), (2r2 − 2r1 + 1, r2 + 2), and (2r2 + 2r1 − 1, r1 + r2 + 2), where r1, r2 are positive integers with r2 > r1 + 1 > 2. Proof See [7]. Lemma 7.

If D is an odd prime and D belongs to the exceptional case (i) of Lemma 6, then D = 17. Proof We now assume that D is an odd prime with D = 22r − 3 · 2r+1 + 1. Then we have (2r−1)2−2r+2=D, (7) (2r+1)2−2r+3=D. (8) If 2∣r, since r ≥ 3, then r ≥ 4, and by (7), we have (2r−1)+2r/2+1=D, (2r−1)−2r/2+1=1. (9) But, by the second equality of (9), we get 1 ≡ (2r − 1) − 2r/2+1 ≡ −1(mod 8), a contradiction. If 2∤r, then from (8) we get (2r+1)+2(r+3)/2=D, (2r+1)−2(r+3)/2=1. (10) Further, by the second equality of (10), we have 2r = 2(r+3)/2, r = 3, and D = 17.

Thus, the lemma is proved. Lemma 8. If D is an odd prime and D belongs to the exceptional case (iii) of Lemma 6, then D = 17. by Lemmas 6–9, (1) has at most two solutions (x, y, z). Thus, the theorem is proved.

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