Last Three Digits of Complex Number Sum and Product Equations

[anemone]

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 $u^4+v^4+x^4+y^4+4uvxy$.

 

[mente oscura]

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

Spoiler

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

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

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

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

Regards.

 

[anemone]

Hello.

Puff. :p

Spoiler

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

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

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

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

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.

 

[mente oscura]

Hello.

My method is: brute force. (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.

 

[mathbalarka]

Pardon for my use of the language Englishman.

Your English is definitely better than my Spanish.