Suppose $a,b,c,d$ are real numbers such that


Show that


My attempt: Using $a+b+c+d=0$, I get

$a(a+b)(a+c)(a+d)= 0 \implies a=0, \text{or}$ $ a(bc+cd+db)+bcd=0$

How can I use $a^7+b^7+c^7+d^7=0$ to prove $a(bc+cd+db)+bcd=0$ ?

Edit:(courtesy @mathguy)

Replacing $d$ by $-(a+b+c)$ we see that the hypothesis is equivalent to

$(a+b+c)^7=a^7+b^7+c^7$, and the conclusion equivalent to $a(a+b)(a+c)(b+c)=0$. The polynomial $(a+b+c)^7-a^7-b^7-c^7$ is divisible by $(a+b)(a+c)(b+c)$, and another irreducible fourth degree symmetric polynomial $P(a,b,c)$ .

Also, $P(0,b,c)= b^4+2b^3c+3b^2c^2+2bc^3+c^4$. It remains to be shown that $P(a,b,c)=0$ when $a=0$

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    $\begingroup$ The two conditions in the hypothesis are equivalent to $(a+b+c)^7 = a^7 + b^7 + c^7$ and the conclusion is equivalent to showing $a(a+b)(a+c)(b+c) = 0$. So we can do away with $d$. Now it is easy to show that $(a+b+c)^7 - a^7 - b^7 - c^7$ is divisible by $(a+b)(b+c)(a+c)$. The quotient, a fourth-degree symmetric polinomial, is irreducible according to Wolfram Alpha. Perhaps it is always $\ge 0$ with equality only when $a = b = c = 0$ for real numbers $a, b, c$; I didn't see how to show that right away, but that would solve the problem. Kind of an unfair problem unless for a math Olympiad...? $\endgroup$
    – mathguy
    Apr 21, 2016 at 18:21
  • $\begingroup$ @mathguy I don't get how the given hypothesis is equivalent to $(a+b+c)^7=a^7+b^7+c^7$. Could you explain in a bit more detail? And it may be from an Olympiad, the book I am following has many problems taken from the olympiads. $\endgroup$ Apr 21, 2016 at 18:45
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    $\begingroup$ Notice that: $$ P(0,b,c)=\color{red}{(b^2+bc+c^2)^2} $$ $\endgroup$ Apr 21, 2016 at 19:36
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    $\begingroup$ And that: $$ P(a,b,c) = (a^2+b^2+c^2+ab+bc+ac)^2 + abc(a+b+c).$$ $\endgroup$ Apr 21, 2016 at 19:38
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    $\begingroup$ This is similar to this old question. $\endgroup$
    – JimmyK4542
    Dec 29, 2016 at 7:12

1 Answer 1


Since $d=-(a+b+c)$ and $d^7=-(a^7+b^7+c^7)$, we obtain $$(a+b+c)^7-a^7-b^7-c^7=0$$ or $$7(a+b)(a+c)(b+c)\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$$ or $$(a+b)(a+c)(a+d)\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0.$$ Thus, it remains to prove that if $\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$ so $a=0$.

We'll prove that $\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)\geq0$.

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$,where $v^2$ can be negative, and $abc=w^3$.

Hence, we need to prove that $f(w^3)\geq0$, where $f$ is a linear function.

But the linear function gets a minimal value for en extremal value of $w^3$.

Since $a$, $b$ and $c$ are real roots of the equation $(x-a)(x-b)(x-c)=0$ or $x^3-3ux^2+3v^2x=w^3$,

we see that the graph of $g(x)=x^3-3ux^2+3v^2x$ and a line $y=w^3$ have three common points

and $w^3$ gets an extremal value, when a line $y=w^3$ is a tangent line to the graph of $g$,

which happens for equality case of two variables.

Since our inequality is symmetric, we can assume $c=b$,

which gives $a^4+4a^3b+11a^2b^2+14ab^3+9b^4\geq0$ or

$$(a+b)^4+ 5(a+b)^2b^2+3b^4\geq0,$$ which is obvious.

The equality occurs for $a=b=0$ and we proved that

if $\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$ so $a=b=c=0$

and we are done!


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