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Difference between revisions of "1985 USAMO Problems/Problem 1"

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==Solution==
 
==Solution==
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{{solution}}
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Lemma: For a positive integer <math>n</math>, <math>1^3+2^3+\cdots +n^3 = (1+2+\cdots +n)^2</math> (Also known as Nicomachus's theorem)
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Proof by induction: The identity holds for <math>1</math>. Suppose the identity holds for a number <math>n</math>. It is well known that the sum of first <math>n</math> positive integers is <math>\frac{n(n+1)}{2} = \frac{n^2+n}{2}</math>. Thus its square is <math>\frac{n^4+2n^3+n^2}{4}</math>. Adding <math>(n+1)^3=n^3+3n^2+3n+1</math> to this we get <math>\frac{n^4+6n^3+13n^2+12n+4}{4}</math>, which can be rewritten as <math>\frac{(n^4+4n^3+6n^2+4n+1)+2(n^3+3n^2+3n+1)+(n^2+2n+1)}{4}</math> This simplifies to <math>\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</math>. The induction is complete.
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Let <math>j</math> be the sum <math>1+2+\cdots 1985</math>, and let <math>k</math> be the sum <math>1^2 + 2^2 + \cdots + 1985^2</math>. Then assign <math>x_i</math> the value <math>ik^4</math> for each <math>i = 1, 2,\cdots 1985</math>. Then:
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<cmath>
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\begin{align*}
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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\\
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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
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\end{align*}
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</cmath>
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Thus, a positive integral solution exists.
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-Circling
  
 
==See Also==
 
==See Also==

Latest revision as of 03:18, 23 October 2022

Problem

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 $x_1,x_2,\cdots,x_{1985}$.

Solution

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.

-Circling

See Also

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

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