Consider this equation: A^x+B^y=C^z Where A,B,C,x,y, and z
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Consider this equation:
A^x+B^y=C^z
Where A,B,C,x,y, and z are positive integers, x,y, and z being more than 2, and the largest common divisor of A,B, and C being 1.
Can this equation exist, and if it is impossible, why?
go to bed wiles
This equation cannot exist. You didn't actually write anything down.
There aren't any solutions if x, y, and z are equal. That's a start.
1^ 5 + 2^3 = 3^2
>>7714977
Not even the answer can have a power of 2
>>7714945
Because of Fermat's Last Theorem.
0^4 + 3^2 = 9^1
Wow this is worth a million bucks. https://en.m.wikipedia.org/wiki/Beal%27s_conjecture
>>7714864
I think when you put that many constraints on it it becomes pointless.
Ever permutation of 2,4,4 for x,y, and z is wrong because of this proof. https://en.m.wikipedia.org/wiki/Proof_by_infinite_descent#Non-solvability_of_r2_.2B_s4_.3D_t4
Every permutation of n,4,4 for x,y, and z is wrong because of this proof. http://www.staff.science.uu.nl/~beuke106/Fermatlectures.pdf
>>7714864
>filename
wat?
>>7714871
this
x,y, and z being 7 or less with A,B, and C being 250,000 or less has no results. x,y, and z being 100 or less with A,B, and C being 10,000 or less has no results.
>>7714864
[math] 2^3 + 2^3 = 2^4 [/math]
Prove by example.
>>7715055
>and the largest common divisor of A,B, and C being 1.
>>7715059
>>7715055
2 goes into 2.
5^3 = 7^2 + [sqrt(76)]^2
/thread
>>7714864
Not possible, I've tried.
The case (x, y, z) = (2, 3, 7) and all its permutations were proven to have only four solutions, none of them involving an even power greater than 2. http://arxiv.org/pdf/math/0508174v1.pdf
14+13=14
>>7715236
Wooops 1^4+1^3=1^4
>>7715238
since when does 1+1 = 1?
1^9+1^7=sqrt(2)^2?
>>7715298
Nevermind. Integers.
>>7715070
Larger than two
>>7715256
Boole.
>>7715256
Boolean expression faggot
The case (x, y, z) = (2, 3, 8) and all its permutations are known to have only three solutions, none of them involving an even power greater than 2 because of this proof http://www.staff.science.uu.nl/~beuke106/Fermatlectures.pdf
>>7714945
Not exactly. For x = y = z = n and a, b, and c all real numbers, there does not exist a set of values that satisfy the function such that n >=3
>a^x
I may be an autistic newfag, but doesn't /sci/ have some sort of plugin or tag similar to the [code] tag on /g/ that makes it so you can write superscripts properly?
>>7715995
>all real numbers
Are you retarded?
0^n + 0^n = 0^n for all n>= 3.
>>7714992
It doesn't, though I understand how someone could think that. Any time you've answered a bunch of questions about something, the questions about that thing become more complex. This is just something a lot of people have asked questions about.>>7715004
Because it's interesting. If that's not enough for you, look up Andrew Wiles and Fermat's Last Theorem. A solution to a question similar to this actually had far reaching consequences in various fields.
>>7714864
Well, it can exist. It may never be valid under the parameters given, but
