# Are there solutions to 4m^(n)=n^(2m) with m,n in Z+

1. Apr 14, 2015

### megatyler30

Hi, thanks for taking the time to read.
To gain insight into $$n^m+n^m+1$$ and when it's prime, I looked at one case where it would be composite.
I equated it to $$(a+1)^2$$ and then substituted out to get $$4m^n=n^{2m}$$ Using mathematica, I was unable to get a solution. So here's my question: are there any solutions to it with integer n and m?

2. Apr 14, 2015

### Staff: Mentor

By inspection, n=1 and m=2 or is that 2m an exponent?

3. Apr 14, 2015

### megatyler30

2m is an exponent. I fixed it.

4. Apr 14, 2015

### Staff: Mentor

There's likely no integer solution. In the simpler case of x^y = y^x the only integer solution is 2 and 4.

My pocket CAS on iOS couldnt find any solutions either but it couldnt solve my easier one either.

5. Apr 14, 2015

### megatyler30

Well that's pretty interesting since no integer solutions would imply that $$x^y+y^x+1$$ will never be a perfect square. Thank you.

Edit: Found m=1 and n=2 is a solution. I wonder if that's the only integer solution.

Last edited: Apr 14, 2015