**Open Access**-
**Authors :**Sanjay Tahiliani -
**Paper ID :**IJERTV9IS040296 -
**Volume & Issue :**Volume 09, Issue 04 (April 2020) -
**Published (First Online):**20-04-2020 -
**ISSN (Online) :**2278-0181 -
**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+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

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)}.

REFERENCES

y y ,

y y ,

W.Ivorra., Sur les equations xp + 2 p= z2 et xp + 2 p= 2z2 Acta Arith.,108:4 (2003),327-338.

J.F.T.Rabago., On an open problem by B.Sroysang, Konuralp J. Math.,1:2 (2013) 30-32.

B.Sryosang., More on Diophantine equation 2x+19y=z2. International journal of pure and applied mathematics,88 (1) 2013, 157-160.

Sanjay Tahiliani., On exponential Diophantine equation 8x+41y =z2.

(communicated).

Y.Yu and X.Li., The exponential Diophantine equation 2x+by=cz, Sci. World J., Vol.2014 ,Article ID 401816,3 pages.