Question

the number of real solutions of the equation (16x^200+1)(y^200+1)=16(xy)^100​

Answer 1

Answer:

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Answer 2

(16x²⁰⁰+1) (y²⁰⁰+1) = 16(x y)¹⁰⁰,x,y∈R

Let x¹⁰⁰ =a and y¹⁰⁰ =b, where a,b are non-negative reals.

So, we have:-

⇒(16a²+1)(b²+1)=16ab

⇒16a²b²+16a²+b²+1= 16ab

Applying AM-GM inequality on the terms 16a2b2,16a2,b2,1 :

(16a²b²+16a²+b²+1 ) ÷ 4 ≥ (4√16a²b²⋅16a²⋅b²⋅1)=4ab

⟹16a²b²+16a²+b²+1 ≥16ab

Since 16a²b2²+16a²+b²+1=16ab, from the equality condition of AM-GM inequality,

16a2b2=16a2=b2=1⟹(a,b)=($\frac{1}{4}$ ,1)

∴(x,y)∈{(100√$\frac{1}{4}$,1),(100√$\frac{1}{4}$,−1),(−100√$\frac{1}{4}$,1),(−100√$\frac{1}{4}$,−1)}

Therefore, there are 4 pairs of reals (x,y) that satisfy the given equation.