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)
number-theory systems-of-equations arithmetic-progressions
 |Â
show 11 more comments
up vote
-2
down vote
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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)
number-theory systems-of-equations arithmetic-progressions
1
What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
â 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.
â 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.
â quasi
Aug 15 at 11:21
yaa thanks for advice
â 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 $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
â quasi
Aug 15 at 11:38
 |Â
show 11 more comments
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)
number-theory systems-of-equations arithmetic-progressions
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)
number-theory systems-of-equations arithmetic-progressions
edited Aug 22 at 17:00
asked Aug 15 at 11:00
Adarsh Kumar
2619
2619
1
What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
â 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.
â 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.
â quasi
Aug 15 at 11:21
yaa thanks for advice
â 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 $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
â quasi
Aug 15 at 11:38
 |Â
show 11 more comments
1
What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
â 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.
â 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.
â quasi
Aug 15 at 11:21
yaa thanks for advice
â 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 $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
â quasi
Aug 15 at 11:38
1
1
What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
â quasi
Aug 15 at 11:08
What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
â 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.
â 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.
â 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.
â 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.
â quasi
Aug 15 at 11:21
yaa thanks for advice
â Adarsh Kumar
Aug 15 at 11:23
yaa thanks for advice
â 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 $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
â 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 $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
â quasi
Aug 15 at 11:38
 |Â
show 11 more comments
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1
What makes you think $$a^x+ (a+d)^x+y= (a+2d)^x+2y$$ ??
â 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.
â 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.
â quasi
Aug 15 at 11:21
yaa thanks for advice
â 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 $$1 + (a+d)^y le (a+d+1)^y < (a^2+4d^2+4ad)^y=(a+2d)^2y$$
â quasi
Aug 15 at 11:38