Complex Multiplication!

Algebra Level 5

a a , b b , c c and d d are non-zero complex numbers that satisfy

a + b + c + d = a 3 + b 3 + c 3 + d 3 = 0 a+b+c+d=a^3+b^3+c^3+d^3=0

The sum of all possible values of ( a + b ) ( a + c ) ( a + d ) (a+b)(a+c)(a+d) can be expressed as p + q i p+qi where p p and q q are real numbers and i i is the imaginary unit that satisfies i 2 = 1 i^2=-1 . What is p + q p+q ?


This problem is inspired by a problem that appeared in ITT - 1994 1994 .


This problem is from the set "Olympiads and Contests Around the World -1". You can see the rest of the problems here .


The answer is 0.

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3 solutions

Mursalin Habib
May 4, 2014

We claim that at least one of ( a + b ) (a+b) , ( a + c ) (a+c) and ( a + d ) (a+d) is equal to zero. So, ( a + b ) ( a + c ) ( a + d ) (a+b)(a+c)(a+d) is always equal to zero.

Now we're going to prove this.

Notice that a 3 + b 3 = ( c 3 + d 3 ) a^3+b^3=-(c^3+d^3)

( a + b ) ( a 2 a b + b 2 ) = ( c + d ) ( c 2 c d + d 2 ) \Rightarrow (a+b)(a^2-ab+b^2)=-(c+d)(c^2-cd+d^2)

As ( c + d ) = a + b -(c+d)=a+b , this equation can be rewritten as:

( a + b ) ( a 2 a b + b 2 ) = ( a + b ) ( c 2 c d + d 2 ) (a+b)(a^2-ab+b^2)=(a+b)(c^2-cd+d^2)

Now if a + b = 0 a+b=0 , we're done. If it's not equal to zero, we can divide both sides of the equation by ( a + b ) (a+b) to get:

a 2 a b + b 2 = c 2 c d + d 2 ( 1 ) a^2-ab+b^2=c^2-cd+d^2 \cdots (1)

( a + b ) 2 3 a b = ( c + d ) 2 3 c d \Rightarrow (a+b)^2-3ab=(c+d)^2-3cd .

Since a + b = ( c + d ) a+b=-(c+d) , ( a + b ) 2 = ( c + d ) 2 (a+b)^2=(c+d)^2 and that means a b = c d ab=cd .

Go back to ( 1 ) (1) and re-write it as:

a 2 + b 2 = c 2 + d 2 a^2+b^2=c^2+d^2

( a 2 c 2 ) + ( b 2 d 2 ) = 0 \Rightarrow (a^2-c^2)+(b^2-d^2)=0

( a + c ) ( a c ) + ( b + d ) ( b d ) = 0 \Rightarrow (a+c)(a-c)+(b+d)(b-d)=0

Substitute b + d b+d with ( a + c ) -(a+c) :

( a + c ) ( a c ) ( a + c ) ( b d ) = 0 (a+c)(a-c)-(a+c)(b-d)=0

( a + c ) ( a c b + d ) = 0 \Rightarrow (a+c)(a-c-b+d)=0

Since c b = a + d -c-b=a+d , we can rewrite the whole thing as:

2 ( a + c ) ( a + d ) = 0 2(a+c)(a+d)=0

This means at least one of ( a + c ) (a+c) and ( a + d ) (a+d) is equal to zero.

So, ( a + b ) ( a + c ) ( a + d ) = 0 (a+b)(a+c)(a+d)=0 and that is the only possible value of that.

That means the sum of all possible values of the given expression is equal to 0 + 0 i 0+0i .

And p + q = 0 p+q=\boxed{0} .


I honestly have no idea what ITT is. I was reading a seminar lecture and came across a problem similar to this.

But good hypothesis and proof nice

Akshay Sant - 7 years ago

Another way to do it is show that the polynomial P ( x ) = ( x a ) ( x b ) ( x c ) ( x d ) P(x)=(x-a)(x-b)(x-c)(x-d) is a polynomial in x 2 x^2 by using vieta's / newton's sums

Shyan Akmal - 7 years ago

we can also prove that one of (a+b),(a+c) and(a+d) is zero,by putting a=1,b=(omega),c=(omega)^2 an d=(omega)^3, where omega is cube root of unity,as they are found to satisfy the equation a + b + c + d =0

Pranav Kilambi - 7 years ago

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Your claim, as Akshay has said, is not true. Even if it were true, that is not a 'proof'. Just because a specific set ( a , b , c , d ) (a, b, c, d) satisfies a certain property doesn't mean that all other sets will do that too. Notice that the problem asks for the sum of all possible values of the product. So, 0 0 may not be the only answer. You have to prove beyond doubt that no matter what a , b , c a, b, c and d d are, the product will always equal zero.

Mursalin Habib - 7 years ago

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@Mursalin Habib yes really... happy following you... plzz follow me too .. and the claim he stated is.. not right as i said...

Akshay Sant - 7 years ago

@Pranav Kilambi how actually as per ur assumptions a +b+c+d = 1+ omega + omegâ^2+ omegâ^3 But omegâ^2+ omega +1=0 and omrgâ^3=1 thus eqn becomes... =0+1 =1 which actually not equal to zero...

Akshay Sant - 7 years ago
Xuming Liang
May 24, 2014

I will be using a lot of sigma summations in this solution, just a heads up.

Interestingly, c y c a = c y c a 3 = 0 \sum_{cyc} a=\sum_{cyc} a^3=0 leads to a lot of other zeros.

It turns out if you expand the desired product you get: a 3 + a 2 ( b + c + d ) + c y c a b c = c y c a b c a^3+a^2(b+c+d)+\sum_{cyc} abc=\sum_{cyc} abc , which is a symmetric sum.

Now consider two constructed products:

( c y c a ) ( c y c a 2 ) = c y c a 3 + c y c a 2 b (\sum_{cyc} a)(\sum_{cyc} a^2)=\sum_{cyc} a^3+\sum_{cyc} a^2b

c y c a 2 b = 0 \Rightarrow \sum_{cyc} a^2b=0

( c y c a ) 3 = c y c a 3 + 3 c y c a 2 b + 6 c y c a b c (\sum_{cyc} a)^3=\sum_{cyc} a^3+3\sum_{cyc} a^2b+6\sum_{cyc} abc

c y c a b c = c y c a 2 b 2 = 0 \Rightarrow \sum_{cyc} abc=-\frac {\sum_{cyc} a^2b}{2}=0 , and that's our answer.

Akshay Sant
May 23, 2014

According to me its bit obivous that two of things a b c d must be equal to 1 and other two equals to -1 thus... let us say that a =b=1 and c=d=-1 to get the equality that a+b+c+d=â^3+b^3+c^3+d^3=0 which satisfies my asumption so directly puting values.. (a+b)(a+c)(a+d)= (1+1)(1+-1)(1+-1) =0 thus p+qi=0+0i thus the answer is 0 ... Thats it

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