Cubes again!
Find the product of all possible integer values of which satisfy the following equation for some
The answer is 22.
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1 solution
Let us write the above Diophantine Equation as:
y 3 − 8 = x 3 + 8 x 2 − 6 x ⇒ ( y − 2 ) ( y 2 + 2 y + 4 ) = x ( x 2 + 8 x − 6 )
Upon observation, ( x , y ) = ( 0 , 2 ) is a trivial solution. Now let:
y − 2 = x ;
y 2 + 2 y + 4 = x 2 + 8 x − 6
which solves as y 2 + 2 y + 4 = ( y − 2 ) 2 + 8 ( y − 2 ) − 6 ⇒ y = 1 1 . Hence ( x , y ) = ( 9 , 1 1 ) is another pair. If we now take:
y − 2 = x 2 + 8 x − 6 ;
y 2 + 2 y + 4 = x
we now obtain y − 2 = ( y 2 + 2 y + 4 ) 2 + 8 ( y 2 + 2 y + 4 ) − 6 ⇒ 0 = y 4 + 4 y 3 + 2 0 y 2 + 3 1 y + 4 4 , which contains two complex conjugate pairs of roots.
There are ultimately two integral pairs ( x , y ) = ( 0 , 2 ) ; ( 9 , 1 1 ) that solve this Diophantine Equation, and our required y-product computes to 2 2 .