MHB Last Three Digits of Complex Number Sum and Product Equations

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The discussion revolves around finding the last three digits of the expression \( u^4 + v^4 + x^4 + y^4 + 4uvxy \) given specific conditions for the complex numbers \( u, v, x, y \). Participants utilize Newton's identities to derive necessary equations from the provided sums and products of the roots. The calculations lead to the conclusion that the last three digits of the desired expression are 560. Various methods, including brute force and systematic substitutions, are discussed to arrive at the solution. The conversation highlights the collaborative nature of problem-solving in mathematics.
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Let $u,v,x,y$ be complex numbers satisfying

$u+v+x+y=42$,

$uv+ux+uy+vx+vy+xy=2013$,

$u^3+v^3+x^3+y^3+uvx+uvy+uxy+vxy=1337$.

Find the last three digits of $u^4+v^4+x^4+y^4+4uvxy$.
 
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anemone said:
Let $u,v,x,y$ be complex numbers satisfying

$u+v+x+y=42$,

$uv+ux+uy+vx+vy+xy=2013$,

$u^3+v^3+x^3+y^3+uvx+uvy+uxy+vxy=1337$.

Find the last three digits of $u^4+v^4+x^4+y^4+4uvxy$.

Hello.

Puff. :p

u^4+v^4+x^4+y^4+4uvxy=

u^3+v^3+x^3+y^3+uvx+uvy+uxy+vxy-

\dfrac{(uv+ux+uy+vx+vy+xy)(u^2+v^2+x^2+y^2)}{u+v+x+y}=

=42*1337-42 \dfrac{2013(42^2-2*2013)}{42}=4609560

Regards.
 
mente oscura said:
Hello.

Puff. :p

u^4+v^4+x^4+y^4+4uvxy=

u^3+v^3+x^3+y^3+uvx+uvy+uxy+vxy-

\dfrac{(uv+ux+uy+vx+vy+xy)(u^2+v^2+x^2+y^2)}{u+v+x+y}=

=42*1337-42 \dfrac{2013(42^2-2*2013)}{42}=4609560

Regards.

Hello mente oscura,

Thanks for participating and your answer is correct.(Clapping)

My method, I believe, is more or less the same as yours.

My solution:

Let $u, v, x, y$ be the roots of a quartic function and $s_n$ represents the sum of the nth powers of those roots. We are asked to evaluate $s_4-uvxy$.

We see that what we have now are
[TABLE="class: grid, width: 700"]
[TR]
[TD]$s_1$[/TD]
[TD]$s_1=42$,[/TD]
[/TR]
[TR]
[TD]$s_2$[/TD]
[TD]$(u+v+x+y)^2=u^2+v^2+x^2+y^2+2(uv+ux+uy+vx+vy+xy)$

$s_1^2=s_2+2(uv+ux+uy+vx+vy+xy)$

$s_2=s_1^2-2(uv+ux+uy+vx+vy+xy)$

$s_2=42^2-2(2013)=-2262$[/TD]
[/TR]
[TR]
[TD]$s_3$[/TD]
[TD]$\tiny(u+v+x+y)^3=u^3+v^3+x^3+y^3+6(uvx+uvy+uxy+vxy)+3((u+v+x+y)(u^2+v^2+x^2+y^2)-(u^3+v^3+x^3+y^3))$

$s_1^3=s_3+6(1337-s_3)+3((s_1)(s_2)-(s_3))$

$s_3=\dfrac{6(1337)+3(s_1)(s_2)-s_1^3}{8}=-43884.75$
[/TD]
[/TR]
[/TABLE]

By applying the values that we have gotten above into the Newton identities gives the quartic equation $f(a)=a^4-42a^3+2013a^2-45221.75a+\text{product of roots}=a^4-42a^3+2013a^2-45221.75a+uvxy$.

By using the Newton Identities again, we get

$s_4(1)+(-42)(s_3)+2013(s_2)+(-45221.75)(s_1)+4uvxy=0$

$s_4+4uvxy=42(s_3)-2013(s_2)+(45221.75)(s_1)$

$s_4+4uvxy=42(-43884.75)-2013(-2262)+(45221.75)(42)=4609560$

Therefore the last three digits of $u^4+v^4+x^4+y^4+uvxy$ is 560.
 
Last edited:
Hello.

My method is: brute force. :mad: (Muscle)

I realized the division:

\dfrac{u^4+v^4+x^4+y^4}{u+v+x+y}=u^3+v^3+x^3+y^3+uvx+uvy+uxy+vxy+R

Being "R" the rest.

I grouped the terms of the rest:

R=-(uv+ux+uy+vx+vy+xy)(u^2+v^2+x^2+y^2)

For being:

u^2+v^2+x^2+y^2=(u+v+x+y)^2-2(uv+ux+uy+vx+vy+xy).

And I realized the necessary substitutions.

Pardon for my use of the language Englishman. How, I do not know it, I use a "on-line" translator and, already we know " the translations that it realizes ".(Rofl)

Regards.
 
mente oscura said:
Pardon for my use of the language Englishman.

Your English is definitely better than my Spanish.
 
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