1985 USAMO Problems/Problem 1

Revision as of 04:18, 23 October 2022 by Circling (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


Determine whether or not there are any positive integral solutions of the simultaneous equations \begin{align*} x_1^2 +x_2^2 +\cdots +x_{1985}^2 & = y^3,\\ x_1^3 +x_2^3 +\cdots +x_{1985}^3 & = z^2 \end{align*} with distinct integers $x_1,x_2,\cdots,x_{1985}$.


Lemma: For a positive integer $n$, $1^3+2^3+\cdots +n^3 = (1+2+\cdots +n)^2$ (Also known as Nicomachus's theorem)

Proof by induction: The identity holds for $1$. Suppose the identity holds for a number $n$. It is well known that the sum of first $n$ positive integers is $\frac{n(n+1)}{2} = \frac{n^2+n}{2}$. Thus its square is $\frac{n^4+2n^3+n^2}{4}$. Adding $(n+1)^3=n^3+3n^2+3n+1$ to this we get $\frac{n^4+6n^3+13n^2+12n+4}{4}$, which can be rewritten as $\frac{(n^4+4n^3+6n^2+4n+1)+2(n^3+3n^2+3n+1)+(n^2+2n+1)}{4}$ This simplifies to $\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4} = ({\frac{(n+1)^2+(n+1)}{2}})^2 = (1+2+\cdots +n+(n+1))^2$. The induction is complete.

Let $j$ be the sum $1+2+\cdots 1985$, and let $k$ be the sum $1^2 + 2^2 + \cdots + 1985^2$. Then assign $x_i$ the value $ik^4$ for each $i = 1, 2,\cdots 1985$. Then: \begin{align*} x_1^2 +x_2^2 +\cdots +x_{1985}^2 & = 1^2k^8 +2^2k^8+\cdots +1985^2k^8 = k^8(1^2+2^2+\cdots +1985^2) = k^9 = {(k^3)}^3\\ x_1^3 +x_2^3 +\cdots +x_{1985}^3 & = 1^3k^{12}+2^3k^{12}+\cdots 1985^3k^{12}=k^{12}(1^3+2^3+\cdots 1985^3) = k^{12}j^2 = ({k^6j})^2 \end{align*}

Thus, a positive integral solution exists.


See Also

1985 USAMO (ProblemsResources)
Preceded by
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png