Okay, That's A Lot Of Integers
x + y + z = 3 x 3 + y 3 + z 3 = 3 Let a be the sum of all possible integer values of x .
Let b be the sum of all possible integer values of y .
Let c be the sum of all possible integer values of z .
Find a + b + c .
Details and Assumptions :
All the possible values of x , y , z are to be summed,not only the possible distinct values of x , y , z .For example, if the solutions were ( 1 , 2 , 3 ) and ( 1 , 2 , 4 ) ,then the answer would be 1 + 1 + 2 + 2 + 3 + 4 = 1 3
The answer is 12.
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2 solutions
Moderator note:
Indeed, from ( 3 − x ) ( 3 − y ) ( 3 − z ) = 8 , we have finitely many integers triplets to check. The condition of ( 3 − x ) + ( 3 − y ) + ( 3 − z ) reduces the number of cases further.
Note: Instead of "any one of them even", you mean that "exactly one of them is even".
Great solution! Here is the proof for the first statement: x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) = ( x + y + z ) ( ( x + y + z ) 2 − 3 ( x y + y z + z x ) ) ⟹ ( x + y + z ) 3 − ( x 3 + y 3 + z 3 ) = 3 ( ( x + y + z ) ( x y + y z + z x ) − x y z ) = 3 ( ( x + y ) ( x y + y z + z x ) + z 2 y + z 2 x ) = 3 ( ( x + y ) ( x y + y z + z x ) + z 2 ( x + y ) ) = 3 ( x + y ) ( z 2 + z x + z y + x y ) = 3 ( x + y ) ( y + z ) ( z + x )
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Here's another approach:
Let
F
(
x
,
y
,
z
)
=
(
x
+
y
+
z
)
3
−
(
x
3
+
y
3
+
z
3
)
.
Observe that
F
(
x
,
−
x
,
z
)
=
(
x
−
x
+
z
)
3
−
(
x
3
−
x
3
+
z
3
)
=
0
.
So
x
+
y
∣
F
(
x
,
y
,
z
)
.
Similarly,
(
x
+
y
)
(
y
+
z
)
(
z
+
x
)
∣
F
(
x
,
y
,
z
)
and so we have
F
(
x
,
y
,
z
)
=
A
(
x
+
y
)
(
y
+
z
)
(
z
+
x
)
.
Finally, from
F
(
1
,
1
,
1
)
=
A
×
2
×
2
×
2
, we conclude that
A
=
3
.
Thanks! I just wrote the conclusion directly and skipped the proof
Note that ( x + y + z ) 3 − ( x 3 + y 3 + z 3 ) = 3 ( x + y ) ( y + z ) ( z + x ) . Substituting the two given equations together gives
( x + y ) ( y + z ) ( x + z ) = 8 ⇒ ( 3 − z ) ( 3 − y ) ( 3 − x ) = 8 ( 1 )
Again, since ( 3 − x ) + ( 3 − y ) + ( 3 − z ) = 6 , which implies that either all x , y , z are even any one of them even.
From ( 1 ) ,
Case 1: All of them are equal to 2, so x = y = z = 1 .
Case 2: Any one of them is 8 and rest are 1: 3 − x = 8 ⇒ x = − 5 , therefore x = − 5 , y = z = 4 . And because x , y , z are interexchangeable, ( x , y , z ) = ( − 5 , 4 , 4 ) , ( 4 , − 5 , 4 ) , ( 4 , 4 , − 5 ) are all the solutions.
Plutting this all together, we have ( x , y , z ) = ( 1 , 1 , 1 ) , ( − 5 , 4 , 4 ) , ( 4 , − 5 , 4 ) , ( 4 , 4 , − 5 ) are all the solutions satisfying the conditions.
So our answer is 3 ( 1 − 5 + 4 + 4 ) = 1 2 .