On exponential Diophantine equation 2x+41y =z2

On exponential Diophantine equation 2x+41y =z2

Sanjay Tahiliani*

*Lecturer, N.K.Bagrodias, Sec 9, Rohini, New Delhi -85

Abstract. In this paper, we used different method and find all solutions of exponential Diophantine equations 2x +41y =z2 ,where x,y,z are non negative integers.

Key words and phrases. Exponential Diophantine equations 2010 Mathematics Subject Classification. 11D09,11D99.

1. INTRODUCTION

   Diophantine analysis deals with various techniques of solving Diophantine equations .If a Diophantine equation has variables as exponents, it is said to be exponential Diophantine equation. In this note, we will find all solutions of exponential Diophantine equation 2x

   2

   2

   +41y =z ,where x,y,z are non negative integers.

2. MAIN RESULTS

Theorem 1. [4] .The only solutions to the Diophantine equation 8x

2

2

+41y =z in non negative integers are given by (x,y,z){(1,0,3),(1,1,7)}.

z

z

y 2

y 2

Theorem 2.[1]. The equation 2x + b = z , b,x,y,zN, gcd(b,z)=1,b>1,x>1,y3 has only solution (b,x,y,z)=(17,7,3,71). Theorem 3.[5] The equation 2x +by =c admits the solution for

50

x>1, y=1, 2z and 2 < 13.

2

2

Theorem 4. The only solutions (x,y,z)N3 to the equation 2x +41y =z are (3,1,7) ,(7,1,13).

Proof. Let x, y, zN. First, we determine the parity of x,y and z. Since x1, then, obviously z is odd. Morever ,its clear that the equation 2x

2 3

2 3

+41y =z may admit a solution (x,y.z)N provided y is odd, otherwise, we will have a contradiction([2]). Suppose now that2x

+41y = z2 holds true for some tuples (x,y,z) where x, y and z are all odd.We consider two cases, case 1 ,where 3x and case 2 is where 3 is not divisible by x.

Case 1. If x=3k, where k is natural, and 2 is not divisible by k, then we have 23k +41y = z2 or equivalently 8k +41y = z2. In view of Theorem 1, the only solution (k,y,z) (x,y.z)N3 is (1,1,7).Hence we get (3,1,7) as only solution to 2x +41y = z2 in N, for x divisible by 3.

2

2

Case 2. Now, suppose that 3 is not divisible by x. First, we suppose that x=1. Then we have 2 +41y = z2.Note that 41y 1(mod4).Hence we get 2 +41y 3(mod4) while z2 1(mod4), a contradiction. Hence x1.Apparently, 41 is not divisible by z because of congruence 2x=0(mod41). So for y3,the only solution we get is (x,y,z)=(7,3,71)(Theorem 2). Now consider the case for y=1.Since z has quadratic exponent, we know that the equation 2x +41y =z may

admit a solution in N such that x<5041 <41. (Theorem 3).

132

Since the bound for x is small, one can effectively use a simple mathematical program to find whether there is any integer x on the interval (1,41) that makes the quantity 2 + 41 an integer.Nevertheles,the values of x that could satisy the equation may be obtained theoretically,and this we show as follows:

2 x

2 x

Rewriting 2x +41y =z as 3((2 +5)/3)=(z+6)(z6),we see that z must be atleast 7.we know that z is odd so z=2c+3 for c2.It follows that 2x

2

2

2 2 x s

+41= (2c+3) = 4c +12c+9,or equivalently, 4c +12c 2 =32.Suppose that c is even, say c=2 m for some s,m N where m is

x

x

2s+2-x 2 s+2-x 5

odd.Then, 2 (2 m +2 3m 1)=2 …….(1).Recall that x>1, 3 is not divisible by x and x is odd, so x must be atleast

1. Thus from (1), we get a contradiction. Hence c cannot be an even integer.Hence, c is odd.For x7 and c is odd, we have 2 x

   +41=4c2+12c+9.

   Hence 2x+32=4c2+12c,which gives 2x-2+8= c2+3c.

   3

   3

   x-5 x- 5 2

   So 2 (2 +1)=c(c+3). Now, c being odd gives c+3=8 or c=5. So 2 =2 and so x=7. Finally this gives the solution

   (x,y,z)=(7,1,13).

   y 2

   y 2

   Hence the final solution for (x,y,z)N3 to 2x +41 =z are(3,1,7)

   ,(7,1,13).

   Remark 1. Further, if x,y,z are non negative, then ,in this equation, x and z can never be 0. Suppose if x=0, the equation becomes 41y= z2

   1 will lead to a contradiction. Indeed, for y and z in N, we have 41(41- 1) = 4141 =(z+1) (z1)=2,where + =y. Evidently, =0 gives 41=3, which is impossible. If z=0, then 2x +41y =0, again impossible for non zero x and y. If y=0, then equation becomes 2x+1=z2 which have a solution (3,0,3)[3].

   3. CONCLUSION

   x

   x

   1 =z

   1 =z

   The all solutions of Diophantine equations 2 +4 y 2 ,where x,y,z are non negative integers is{(3,0,3), (3,1,7)

   ,(7,1,13)}.

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   y y ,

   y y ,

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   5. (communicated).

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