Consider this equation: A^x+B^y=C^z Where A,B,C,x,y, and z

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

>>7714983
>>7714977
Are you guys even able to read?

>>7714998
See >>7714992

Lots of "functions," (infinitely many) have no solutions over R or C. Big fucking deal if this one doesn't.

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.

File: doh.gif (14KB, 581x600px) Image search: [Google]

14KB, 581x600px

>>7715058

File: 1449538579709.jpg (46KB, 473x500px) Image search: [Google]

46KB, 473x500px

>>7715059

>>7715055
2 goes into 2.

5^3 = 7^2 + [sqrt(76)]^2

/thread

File: hisv.jpg (241KB, 1155x897px) Image search: [Google]

241KB, 1155x897px

>>7715070

>>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?

>>7715236
>>7715238
two wrongs do not make a right

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

>>7715004
>Big fucking deal
That's right, Cletus:
>>7714988