Help me with solving this equatation please!

So as you see this is a fucked up equatation, that states:

2*x^x = y^y + z^z, while x,y,z are positive integers.

We have to find all the possible solutions, and give an answer stating while are these the only possible solutions.

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קוסגרוב

Trivial solution : x=y=z=0
Easy solution : x=y=z=1

I don't know if there are more and if there isn't, I can't say why.

We start off by noticing that x and y are either both even or both odd, if precisely one of them is even then x^x + y^y wouldnt be even.

Now assume that both x and y are even, this implies that the right hand side has an even multiplicity of 2 (its the sum of two numbers with even multiplicity of 2).

2 either has even multiplicity or a multiplicity of 0 in x^x (if x is odd its 0 and if x is even i hope you can see why the multiplicity is even). So 2 will always have odd multiplicity in 2*x^x, a contradiction.

y and z must therefore both be odd, that's the best I can do buddy.

I'd generate the first 100 solutions using a simulation, then try to find a relationship between the solutions.

Your "trivial solution" doesn't make sense. What's 0^0?

I'm assuming we're talking integers here.

When x=y=z, there is only one solution (this reduces to fermat's last theorem). Wild guess: for all x,y,z there is only the the solution x=y=z=1 for the integers, since you can just consider
x^k * x^n = y^j * y^n + z^l * z^n
where k+n = x, j+n = y and l+n = z

When in doubt, just take
f(x, y, z) = 2*x^x - y^y + z^z = 0
and differentiate if you're not only considering integer solutions.

>What's 0^0

1 you mong

>What's 0^0?
it's 1, do you even math?

>0^0 is a trivial solution
k

uh huh, alright
so... what is the limit of x^(1/ln(x)) as x -> 0+?