Solving Difficult Questions: Find x & y

[sambarbarian]

Hey , guys . This problem was asked in an olympiard and i don't have a clue of how to solve it.

1) If x + y = 5 and x^y + y^x = 17 .. find all possible values of x and y .

2) 16^2x+y + 16^y+2x = 1 , find all possible values of x and y .

In the first question i tried to convert the whole equation into y but that didn't work.

In the second question i got to :- 16^x + 16^y = 16^-x-y .

 

[jedishrfu]

I assuming the problem means integer solutions

x + y = 5 means you should sub in 0,5 1,4 2,3 3,2 4,1 5,0 choices

then check more exotic ones -1, 6 ... but they would get dropped since only one term would be a fraction

2,3 and 3,2 solutions work for the first.

Perhaps you can use the same kind of analysis on the second one as well...

 

[Fightfish]

For first question, I believe that you have to show/explain that the only allowed solutions are ordered pairs of integers. Then the problem becomes trivial.

 

[sambarbarian]

but we were'nt meant to do it by hit and trial ( it was specified )

 

[Ray Vickson]

Hey , guys . This problem was asked in an olympiard and i don't have a clue of how to solve it.

1) If x + y = 5 and x^y + y^x = 17 .. find all possible values of x and y .

2) 16^2x+y + 16^y+2x = 1 , find all possible values of x and y .

In the first question i tried to convert the whole equation into y but that didn't work.

In the second question i got to :- 16^x + 16^y = 16^-x-y .

Question 2 is very unclear. Do you mean
16^2 x + y + 16^y + 2x = 1? That is what is meant by what you wrote. Do you mean
16^{2x} + y + 16^y + 2x = 1,
or do you mean
16^{2x+y} + 16^{y + 2x} = 1? If you mean this last one USE PARENTHESES, like this: 16^(2x+y) + 16^(y+2x) = 1.

 

[jedishrfu]

but we were'nt meant to do it by hit and trial ( it was specified )

So you do it by analysis like I kind of did. One thing about these problems is that you have to think outside of the box meaning sometimes you use you intuition and not strictly follow the rules or perceived rules.

What got me interested in this problem was its similarity to solutions to the x^y = y^x equation that my friend who many many years ago was fascinated by (I have no idea why but it was his thing to do) and he was a US MAA champion.

 

[Saitama]

I guess transforming the equation might help. Write y=5-x
Substitute y in the second equation, you get:
x^{(5-x)}+(5-x)^x=17

Now you can put some constraints on the values of x. x cannot be greater than 5 and less than 0 because that would result in a fraction. Now substitute different values of x from 0 to 5.

Hope that helped.

 

[sambarbarian]

ok , i got the first one , and the second one was 16^(2x+y) + 16^(y+2x) = 1.

 

[Mentallic]

ok , i got the first one , and the second one was 16^(2x+y) + 16^(y+2x) = 1.

Are you sure it's not supposed to be

16^{2x+y}+16^{2y+x}=1

Because what you've written is simply

2\cdot 16^{2x+y}=1