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2 solutions
Easier than I expect! Though, I am not sure if I can consider mine as the perfect solution.
If you already know that the solutions are ( 4 , 2 ) from the inspiration, then by symmetry and exponentiation, we have ( − 4 , − 2 ) . Likewise, we also have ( 2 , 4 ) and ( − 2 , − 4 ) by the same reasoning. It's like obtaining the same values for y = 2 x .
Of course, there are no working solutions with opposite signs, like x = 1 and y = − 1 .
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I forgot to mention that there are no real solutions for opposite-signed solutions.
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Yes, that's a case that most people would not take care of.
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@Calvin Lin – I'm very surprised that my simple approach works for this Level 5 problem. :)
Aren't there infinite real solutions? As an explicit example, y = 3 , x = 3 3 satisfies the given system.
what is the complete solution like how do we do it algebraically
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If the choices were to show at least 2, say 2, 4, 6 and so on, the problem would be difficult to solve. Viewers would be stricken to see possible number of solutions higher than 4. However, I believe Calvin thinks that people would not consider multiple solutions for certain factor integer, say x = 2 y and y = 2 x , where k is either 2 1 or 2 .
Otherwise, if I have figured out 4 integer solutions, and that is the highest value from the choices, the problem would be very easy to answer, and there is no need to spot-check other solutions. It will end up not being a "trick" problem.
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Thanks Michael
When restricted to integer solutions, a simple way to show that there is only this "pair" of solutions is to consider the graph of x x 1 which has a maximum at e . If you don't want to use calculus, you can inequalities to prove that the signs work the way we desire them to.
When relaxed to positive real numbers, there are infinitely many solutions. There is a way to characterize all of them. Once again, look at x x 1 .
I'm avoiding talking negative real numbers because these expressions become multi-valued functions instead.
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@Calvin Lin – Thanks Calvin sir
@Calvin Lin – I find this site very interesting to look at, even though it covers rational integers. In my opinion, I find this problem to lie on the calculus side more than the algebra side.
@Calvin Lin – Thanks Calvin! That works!
You tricking but i'm true Haha(:
Firstly, both must have the same sign because if they have opposite signs , then one side will become non integral and the other side integral.
Now clearly both x and y share the same prime factors and the power of all the prime factors are in the same ratio. Then WLOG x ≥ y ,
Let x = y k , where k ∈ Z + , k > 1
The equation becomes k = y k − 1 . An easy induction shows that if k < 3 and y > 1 , then y k − 1 > k . This leads to only one solution ( k , y ) = ( 2 , 2 ) which corresponds to ( x , y ) = ( 4 , 2 )
Now since the equations are symmetric , we have the solutions ( 4 , 2 ) ; ( 2 , 4 ) .
And taking the case when both x and y are negative , it will lead to a similar equation to give the solutions ( − 2 , − 4 ) ; ( − 4 , − 2 ) .
Altogether 4 solutions.
"Share the same prime factors" doesn't imply that " x = y k , k ∈ Z .
For example, x = 1 8 , y = 1 2 share the same prime factors of 2 and 3. But we don't have x = y k .
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Sir , I forgot to mention that the powers of all the prime factors must be in the same ratio.
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That should help patch the hole. Can you add that in?
After that, the "which clearly only has one solution" should be better explained.
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@Calvin Lin – Sir , are any further edits required?
[This is not a complete solution.]
Most people are familiar with the solutions ( 4 , 2 ) and ( 2 , 4 ) .
However, there are additional solutions of ( − 2 , − 4 ) and − 4 , − 2 ) .