 Open Access
 Authors : Sanjay Tahiliani
 Paper ID : IJERTV9IS040296
 Volume & Issue : Volume 09, Issue 04 (April 2020)
 Published (First Online): 20042020
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
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.

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.

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+2x 2 s+2x 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

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 2x2+8= c2+3c.
3
3
x5 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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