1120 and 1
can be a solution.
Got this one from goiit.. (Nice that i went there and browsed a couple of posts :)
21 = x^y + y^x
Find the values of x and y . (Restrict to the integral solutions only)
Discuss the other cases too...
-
UP 0 DOWN 0 1 9
9 Answers
62Is that the only solution.. and can we think for it in a more general way?
3is there ne soln not by observation :P,bcos the function xy and its inverse was complicated to me.
62No I dont think there is another solution not by observation..
BUt by graphs ther is a way if you can think nicely
11bhaiya only integral solutions is asked ..dont think there is any other integral solution.
1x^y<21 x≠1 y≠1
so permissible values of
x y
2 2,3,4
3 2,
4 2
we see none of these satisfy
hence 20,1 and 1,20 are only soln
for x^y+y^x=8
proceeding similarly we get
1,8 8,1 and 2,2
11Alternately we can do something of this kind ....
(1,20) & (20,1) are the obvious ones....Other than those....if we have both x\ge 1\\ & y\ge 1\\ then :-
y^x\ge 4 and x^y\ge x^2 so 21\ge x^2+4
Meaning probable x can be (2,3,4) Easy to see none satisfy the eqn....
how ?
Like this :-
Suppose x=4, then y^4-21=-4^y ,meaning y can be at max 2.....checking like this, we confirm no other solutions.....
BTW check the following link!