Series of trial-and-error questions: Find all the unknowns for: 1. x^y=y^x 2. x^y=y^x+1 where all the unknowns are integers.

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Series of trial-and-error questions: Find all the unknowns for: 1. x^y=y^x 2. x^y=y^x+1 where all the unknowns are integers.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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?????????!!!!!!!!
i guess \[y=x\] might work for first one
could be...

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Other answers:

let's say they aren't, some actual answers please...
what is wrong with \[3^3=3^3\]?
2 and 4 works for the first one.
Nothing! Pratu got it right :) (Well you too...)
Sorry, I thought it was a system of two equations.
oh.
ooooooooooooooh. i was wondering...
I agree with satellite on the first one, the solution of the first one is all integers (x,y) such that x=y. For the second one, we have (x,y)=(2,1) is a solution.
(3,2) is another solution.
pratu had \[2^4=4^2\] so that works as well. my guess is there are no others, but that is a guess
There are no others.
no proof tho. just know that for SURE :)
Mr. Math, you're missing one more answer. (which i know)
(solution)
ans. in half a min.
(1,0).
YAY!
Actually any solution in the form (x,0), x is a positive integer.
Can we do this for +2, +3, +4 etc.
Yes.

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