MHB Prove (p+r)(p+s)(p+t)(p+u)=(q+r)(q+s)(q+t)(q+u)

[anemone]

If $p+q+r+s+t+u=0$ and $p^3+q^3+r^3+s^3+t^3+u^3=0$, prove that

$(p+r)(p+s)(p+t)(p+u)=(q+r)(q+s)(q+t)(q+u)$.

 

[Opalg]

If $p+q+r+s+t+u=0$ and $p^3+q^3+r^3+s^3+t^3+u^3=0$, prove that

$(p+r)(p+s)(p+t)(p+u)=(q+r)(q+s)(q+t)(q+u)$.

[sp]
Let $f(x) = (x+r)(x+s)(x+t)(x+u) = x^4 + ax^3 + bx^2 + cx + d$, where $a = r+s+t+u$, $b = rs+rt+ru+st+su+tu$, $c = stu + rtu + rsu + rst$ and $d = rstu$. By one of Newton's identities, $r^3+s^3+t^3+u^3 = a^3 - 3ab + 3c$.

So we are told that $p+q = -a$, and $p^3+q^3 = -a^3 + 3ab - 3c$. It follows that $-a^3 = (p+q)^3 = p^3 + q^3 + 3pq(p+q) = -a^3 + 3ab - 3c - 3apq$, from which $$ab - c - apq = 0.\quad (*)$$

We want to show that $f(p) = f(q)$. In fact, $$\begin{aligned} f(p) - f(q) &= p^4 - q^4 + a(p^3 - q^3) + b(p^2-q^2) + c(p-q) \\ &= (p-q)\bigl( p^3 + p^2q + pq^2 + q^3 + a(p^2 + pq + q^2) + b(p+q) + c \bigr) \\ &= (p-q)\bigl( (p+q)^3 - 2pq(p+q) + a((p+q)^2 - pq) + b(p+q) + c \bigr) \\ &= (p-q)\bigl( -a^3 + 2apq + a(a^2 - pq) -ba + c \bigr) \\ &= (p-q)( apq - ab + c) = 0, \quad \text{by }(*). \end{aligned}$$[/sp]

 

[anemone]

Very well done, Opalg, and welcome back home! :)

 

[anemone]

Solution of other:

Spoiler

Let $a_k$ be the sum of the products of the four letters $r,\,s,\,t,\,u$, taken $k$ at a time.

Then $a_1=r+s+t+u=-(p+q)$.

Also,

$\begin{align*}a_1(a_1^2-3a_2)&=(r^3+s^3+t^3+u^3)-3a_3\\&=-(p^3+q^3)-3a_3\end{align*}$

Therefore,

$\begin{align*}a_3&=\dfrac{(p+q)^3-(p^3+q^3)}{3}-a_2(p+q)\\&=(p+q)(pq-a_2)\end{align*}$

Hence,

$\begin{align*}(p+r)(p+s)(p+t)(p+u)&=(q+r)(q+s)(q+t)(q+u)\\&=p^4-p^3(p+q)+p^2a_2+p(p+q)(pq-a_2)+a_4\\&=p^2q^2-pqa_2+a_4\\&=(q+r)(q+s)(q+t)(q+u)\end{align*}$