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Number Theory Level 4

What is the number of real solutions ( x , y ) (x,y) of the equation ( 16 x 200 + 1 ) ( y 200 + 1 ) = 16 ( x y ) 100 ? (16x^{200}+1)(y^{200}+1)=16(xy)^{100}?

Source : NMTC 2015 Inter Level.


The answer is 4.

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Surya Prakash
Aug 25, 2015

Let a = 4 x 100 a=4x^{100} and b = y 100 b=y^{100} . It means that a ≥ 0 a \geq 0 and b ≥ 0 b \geq 0 .

So, the given equation becomes

( a 2 + 1 ) ( b 2 + 1 ) = 4 a b (a^2 + 1)(b^2 + 1) = 4ab ( a b − 1 ) 2 + ( a − b ) 2 = 0 (ab-1)^2 + (a-b)^2 = 0

So, a b = 1 ab =1 and a = b a= b .

So, a = 1 a=1 and b = 1 b=1 . It implies that x = ± ( 1 4 ) 1 100 x= \pm (\dfrac{1}{4})^{\dfrac{1}{100}} and y = ± 1 y= \pm 1 .

Therfore, ( x , y ) = { ( 4 − 0.01 , 1 ) , ( 4 − 0.01 , − 1 ) , ( − 4 − 0.01 , 1 ) , ( − 4 − 0.01 , − 1 ) } (x,y) = \left\{(4^{-0.01},1),(4^{-0.01},-1),(-4^{-0.01},1),(-4^{-0.01},-1)\right\} .

So, there are four possibilities of real solutions ( x , y ) (x,y) .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Interesting way to view the factorization.

There is another "more direct" way that uses inequalities, and exploits the equality condition. It is similar to how you used the trivial inequality to establish that the squares must both be 0.

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