Diophantine equation...
x 3 ( x + y ) + ( y 2 + 1 ) ( x 2 + x y + y 2 ) = 0
Find number of ordered pairs of ( x , y ) .
The answer is 1.
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1 solution
Where does the right half of your first equation come from?
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On multiplying, we will get x 4 + x 3 y + x 2 y 2 + x y 3 + y 4 + x 2 + x y + y 2 = 0 On substituting x=y, 5 x 4 + 3 x 2 = 0 ⟹ x = 0 ⟹ y = 0
How come do you say that f(x)=f(y) ?
Assume x=y for a start, x 3 ( x + y ) + ( y 2 + 1 ) ( x 2 + x y + y 2 ) = 0 ⇒ 2 x 4 + 3 x 2 ( x 2 + 1 ) = 0 ⇒ x = y = 0
Now, If x = y , we can multiply ( x − y ) to both sides of equation giving
x 5 + x 3 = y 5 + y 3 Let f ( x ) = x 5 + x 3 ⇒ f ′ ( x ) = 5 x 4 + 3 x 2 > 0
Hence, f(x) is strictly increasing. f ( x ) = f ( y ) ⇒ x = y which contradicts our assumption.
Therefore ( x , y ) = ( 0 , 0 ) (one solution)
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