Hello,
I am stuck on this question:
Does the expression have a finite or infinite positive pair of integer solutions that satisfy the relation? Prove either way.
The only example I can think of is
I am assuming x doesn't equal y.
If x=1 and y=0, that is a solution. As for proof, maybe contradicition?
So assume that x^y=x+y for an infinite number of paired values belonging to the set N. With x and y being different, well one thing you could say is since the numbers x=3 and y=4 don't work, there has to be a limit to the amount of responses and thus saying infinite pairs of soloutions work is wrong. Alternitvely, an odd + an odd= an even, but given how exponents work, an odd times an odd gives an odd, therefore x and y can't BOTH belong to odd positive integers, an even and an even add to an even, and an even times an even also makes an even so if both x and y are divisible by 2 it works. Adding an odd and an even make an odd number, and if x is the odd number and y is the even it works, but if y is odd and x is even it does not, so it only works when x is even.
Therefore, only when both x and y are divisible by 2 can this equation be true, or if x is odd and y is even. As a result, I would use that to say that there is a finite amount of positive integers. But I am probably wrong so sorry.