Problem of Arithmetic Equation

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I was trying to prove the Beal's Conjecture by using Arithmetic Progression but suddenly I got a new idea but after getting the result of that idea I can't able to get the solution of the equation I got.



Beal's Conjecture states that



$$A^x + B^y = C^z$$



where $A, B, C, x, y, z in mathbb Z_+$, and $x, y, z> 2$



then A,B,and C have a common prime factor



Now what I think is:

$$a^x + (a + d)^x+y = (a+2d)^x+2y tag1$$
I truly don't know why I think these values in Arithmetic Progression. I just choose them and got this above relation



and then taking both sides to limit $x to 0$ in eq(1) because I wanted to know if the value of x approach to zero then where the value of y approach, then I got an equation whose solution is still unknown to me. The equation is
$$1 + (a+d)^y = (a+2d)^2y, tag2$$



Now the main question is that is there any value of $a,d,x$ and $y$ exist which satisfies equation(1)







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  • 1




    What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
    – quasi
    Aug 15 at 11:08











  • a week ago i completed modular arithmetic so i think that i will take these values in Arithmetic Progression. I actually wanted to see the beauty of mathematics, so i always create special cases and then try to understand them.
    – Adarsh Kumar
    Aug 15 at 11:12






  • 3




    Your statement of Beal's Conjecture needs some edits. Presumably you meant $C^z$, not $C^y$. But more importantly, Beal's conjecture claims that if the equation holds, then $gcd(A,B,C) > 1$. Thus, you haven't correctly stated the conjecture.
    – quasi
    Aug 15 at 11:21











  • yaa thanks for advice
    – Adarsh Kumar
    Aug 15 at 11:23






  • 1




    As regards positive integer solutions to the equation $$1 + (a+d)^y =(a+2d)^2y$$ there are no solutions, since for real numbers $a,d,yge 1$, we have $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
    – quasi
    Aug 15 at 11:38














up vote
-2
down vote

favorite












I was trying to prove the Beal's Conjecture by using Arithmetic Progression but suddenly I got a new idea but after getting the result of that idea I can't able to get the solution of the equation I got.



Beal's Conjecture states that



$$A^x + B^y = C^z$$



where $A, B, C, x, y, z in mathbb Z_+$, and $x, y, z> 2$



then A,B,and C have a common prime factor



Now what I think is:

$$a^x + (a + d)^x+y = (a+2d)^x+2y tag1$$
I truly don't know why I think these values in Arithmetic Progression. I just choose them and got this above relation



and then taking both sides to limit $x to 0$ in eq(1) because I wanted to know if the value of x approach to zero then where the value of y approach, then I got an equation whose solution is still unknown to me. The equation is
$$1 + (a+d)^y = (a+2d)^2y, tag2$$



Now the main question is that is there any value of $a,d,x$ and $y$ exist which satisfies equation(1)







share|cite|improve this question


















  • 1




    What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
    – quasi
    Aug 15 at 11:08











  • a week ago i completed modular arithmetic so i think that i will take these values in Arithmetic Progression. I actually wanted to see the beauty of mathematics, so i always create special cases and then try to understand them.
    – Adarsh Kumar
    Aug 15 at 11:12






  • 3




    Your statement of Beal's Conjecture needs some edits. Presumably you meant $C^z$, not $C^y$. But more importantly, Beal's conjecture claims that if the equation holds, then $gcd(A,B,C) > 1$. Thus, you haven't correctly stated the conjecture.
    – quasi
    Aug 15 at 11:21











  • yaa thanks for advice
    – Adarsh Kumar
    Aug 15 at 11:23






  • 1




    As regards positive integer solutions to the equation $$1 + (a+d)^y =(a+2d)^2y$$ there are no solutions, since for real numbers $a,d,yge 1$, we have $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
    – quasi
    Aug 15 at 11:38












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











I was trying to prove the Beal's Conjecture by using Arithmetic Progression but suddenly I got a new idea but after getting the result of that idea I can't able to get the solution of the equation I got.



Beal's Conjecture states that



$$A^x + B^y = C^z$$



where $A, B, C, x, y, z in mathbb Z_+$, and $x, y, z> 2$



then A,B,and C have a common prime factor



Now what I think is:

$$a^x + (a + d)^x+y = (a+2d)^x+2y tag1$$
I truly don't know why I think these values in Arithmetic Progression. I just choose them and got this above relation



and then taking both sides to limit $x to 0$ in eq(1) because I wanted to know if the value of x approach to zero then where the value of y approach, then I got an equation whose solution is still unknown to me. The equation is
$$1 + (a+d)^y = (a+2d)^2y, tag2$$



Now the main question is that is there any value of $a,d,x$ and $y$ exist which satisfies equation(1)







share|cite|improve this question














I was trying to prove the Beal's Conjecture by using Arithmetic Progression but suddenly I got a new idea but after getting the result of that idea I can't able to get the solution of the equation I got.



Beal's Conjecture states that



$$A^x + B^y = C^z$$



where $A, B, C, x, y, z in mathbb Z_+$, and $x, y, z> 2$



then A,B,and C have a common prime factor



Now what I think is:

$$a^x + (a + d)^x+y = (a+2d)^x+2y tag1$$
I truly don't know why I think these values in Arithmetic Progression. I just choose them and got this above relation



and then taking both sides to limit $x to 0$ in eq(1) because I wanted to know if the value of x approach to zero then where the value of y approach, then I got an equation whose solution is still unknown to me. The equation is
$$1 + (a+d)^y = (a+2d)^2y, tag2$$



Now the main question is that is there any value of $a,d,x$ and $y$ exist which satisfies equation(1)









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 22 at 17:00

























asked Aug 15 at 11:00









Adarsh Kumar

2619




2619







  • 1




    What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
    – quasi
    Aug 15 at 11:08











  • a week ago i completed modular arithmetic so i think that i will take these values in Arithmetic Progression. I actually wanted to see the beauty of mathematics, so i always create special cases and then try to understand them.
    – Adarsh Kumar
    Aug 15 at 11:12






  • 3




    Your statement of Beal's Conjecture needs some edits. Presumably you meant $C^z$, not $C^y$. But more importantly, Beal's conjecture claims that if the equation holds, then $gcd(A,B,C) > 1$. Thus, you haven't correctly stated the conjecture.
    – quasi
    Aug 15 at 11:21











  • yaa thanks for advice
    – Adarsh Kumar
    Aug 15 at 11:23






  • 1




    As regards positive integer solutions to the equation $$1 + (a+d)^y =(a+2d)^2y$$ there are no solutions, since for real numbers $a,d,yge 1$, we have $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
    – quasi
    Aug 15 at 11:38












  • 1




    What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
    – quasi
    Aug 15 at 11:08











  • a week ago i completed modular arithmetic so i think that i will take these values in Arithmetic Progression. I actually wanted to see the beauty of mathematics, so i always create special cases and then try to understand them.
    – Adarsh Kumar
    Aug 15 at 11:12






  • 3




    Your statement of Beal's Conjecture needs some edits. Presumably you meant $C^z$, not $C^y$. But more importantly, Beal's conjecture claims that if the equation holds, then $gcd(A,B,C) > 1$. Thus, you haven't correctly stated the conjecture.
    – quasi
    Aug 15 at 11:21











  • yaa thanks for advice
    – Adarsh Kumar
    Aug 15 at 11:23






  • 1




    As regards positive integer solutions to the equation $$1 + (a+d)^y =(a+2d)^2y$$ there are no solutions, since for real numbers $a,d,yge 1$, we have $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
    – quasi
    Aug 15 at 11:38







1




1




What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
– quasi
Aug 15 at 11:08





What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
– quasi
Aug 15 at 11:08













a week ago i completed modular arithmetic so i think that i will take these values in Arithmetic Progression. I actually wanted to see the beauty of mathematics, so i always create special cases and then try to understand them.
– Adarsh Kumar
Aug 15 at 11:12




a week ago i completed modular arithmetic so i think that i will take these values in Arithmetic Progression. I actually wanted to see the beauty of mathematics, so i always create special cases and then try to understand them.
– Adarsh Kumar
Aug 15 at 11:12




3




3




Your statement of Beal's Conjecture needs some edits. Presumably you meant $C^z$, not $C^y$. But more importantly, Beal's conjecture claims that if the equation holds, then $gcd(A,B,C) > 1$. Thus, you haven't correctly stated the conjecture.
– quasi
Aug 15 at 11:21





Your statement of Beal's Conjecture needs some edits. Presumably you meant $C^z$, not $C^y$. But more importantly, Beal's conjecture claims that if the equation holds, then $gcd(A,B,C) > 1$. Thus, you haven't correctly stated the conjecture.
– quasi
Aug 15 at 11:21













yaa thanks for advice
– Adarsh Kumar
Aug 15 at 11:23




yaa thanks for advice
– Adarsh Kumar
Aug 15 at 11:23




1




1




As regards positive integer solutions to the equation $$1 + (a+d)^y =(a+2d)^2y$$ there are no solutions, since for real numbers $a,d,yge 1$, we have $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
– quasi
Aug 15 at 11:38




As regards positive integer solutions to the equation $$1 + (a+d)^y =(a+2d)^2y$$ there are no solutions, since for real numbers $a,d,yge 1$, we have $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
– quasi
Aug 15 at 11:38















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