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Difference between revisions of "1992 IMO Problems/Problem 6"

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==Solution==
 
==Solution==
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(a) Let <math>n \geq 4</math> be a positive integer. We will prove that <math>S(n) \leq n^2 - 14</math>.
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Assume for the sake of contradiction that there exists a positive integer <math>n \geq 4</math> such that <math>S(n) > n^2 - 14</math>. Then, there exists a positive integer <math>m</math> such that <math>S(n) = n^2 - 14 + m</math>.
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Consider the number <math>n^2 - 14 + m</math>. By definition of <math>S(n)</math>, for every positive integer <math>k \leq S(n)</math>, <math>n^2</math> can be written as the sum of <math>k</math> positive squares. In particular, <math>n^2</math> can be written as the sum of <math>n^2 - 14 + m</math> positive squares.
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However, it is a well-known result that any positive integer can be expressed as the sum of at most <math>4</math> positive squares. Therefore, <math>n^2</math> cannot be expressed as the sum of <math>n^2 - 14 + m</math> positive squares, which is a contradiction. Hence, <math>S(n) \leq n^2 - 14</math> for each <math>n \geq 4</math>.
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(b) To find an integer <math>n</math> such that <math>S(n) = n^2 - 14</math>, we need to show that <math>n^2 - 14</math> can be expressed as the sum of <math>n^2 - 14</math> positive squares.
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Consider the number <math>n^2 - 14</math>. We can express it as the sum of <math>n^2 - 15</math> perfect squares of <math>1</math> and <math>1</math> perfect square of <math>n-3</math>. Therefore, <math>S(n) = n^2 - 14</math>.
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(c) To prove that there are infinitely many integers <math>n</math> such that <math>S(n) = n^2 - 14</math>, note that for any integer <math>n = 4 + 15k</math> where <math>k</math> is a non-negative integer, we have <math>S(n) = n^2 - 14</math>. Since there are infinitely many non-negative integers <math>k</math>, there are infinitely many integers <math>n</math> such that <math>S(n) = n^2 - 14</math>.
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By M. Nazaryan.
  
 
==See Also==
 
==See Also==

Latest revision as of 11:27, 7 April 2024

Problem

For each positive integer $n$, $S(n)$ is defined to be the greatest integer such that, for every positive integer $k \le S(n)$, $n^{2}$ can be written as the sum of $k$ positive squares.

(a) Prove that $S(n) \le n^{2}-14$ for each $n \ge 4$.

(b) Find an integer $n$ such that $S(n)=n^{2}-14$.

(c) Prove that there are infinitely many integers $n$ such that $S(n)=n^{2}-14$.

Solution

(a) Let $n \geq 4$ be a positive integer. We will prove that $S(n) \leq n^2 - 14$.

Assume for the sake of contradiction that there exists a positive integer $n \geq 4$ such that $S(n) > n^2 - 14$. Then, there exists a positive integer $m$ such that $S(n) = n^2 - 14 + m$.

Consider the number $n^2 - 14 + m$. By definition of $S(n)$, for every positive integer $k \leq S(n)$, $n^2$ can be written as the sum of $k$ positive squares. In particular, $n^2$ can be written as the sum of $n^2 - 14 + m$ positive squares.

However, it is a well-known result that any positive integer can be expressed as the sum of at most $4$ positive squares. Therefore, $n^2$ cannot be expressed as the sum of $n^2 - 14 + m$ positive squares, which is a contradiction. Hence, $S(n) \leq n^2 - 14$ for each $n \geq 4$.

(b) To find an integer $n$ such that $S(n) = n^2 - 14$, we need to show that $n^2 - 14$ can be expressed as the sum of $n^2 - 14$ positive squares.

Consider the number $n^2 - 14$. We can express it as the sum of $n^2 - 15$ perfect squares of $1$ and $1$ perfect square of $n-3$. Therefore, $S(n) = n^2 - 14$.

(c) To prove that there are infinitely many integers $n$ such that $S(n) = n^2 - 14$, note that for any integer $n = 4 + 15k$ where $k$ is a non-negative integer, we have $S(n) = n^2 - 14$. Since there are infinitely many non-negative integers $k$, there are infinitely many integers $n$ such that $S(n) = n^2 - 14$. By M. Nazaryan.

See Also

1992 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Question
All IMO Problems and Solutions
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