Difference between revisions of "1992 IMO Problems/Problem 6"
(→Solution) |
|||
Line 10: | Line 10: | ||
==Solution== | ==Solution== | ||
− | + | (a) Let <math>n \geq 4</math> be a positive integer. We will prove that <math>S(n) \leq n^2 - 14</math>. | |
+ | |||
+ | 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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | (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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | (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>. | ||
==See Also== | ==See Also== |
Revision as of 12:07, 7 April 2024
Problem
For each positive integer ,
is defined to be the greatest integer such that, for every positive integer
,
can be written as the sum of
positive squares.
(a) Prove that for each
.
(b) Find an integer such that
.
(c) Prove that there are infinitely many integers such that
.
Solution
(a) Let be a positive integer. We will prove that
.
Assume for the sake of contradiction that there exists a positive integer such that
. Then, there exists a positive integer
such that
.
Consider the number . By definition of
, for every positive integer
,
can be written as the sum of
positive squares. In particular,
can be written as the sum of
positive squares.
However, it is a well-known result that any positive integer can be expressed as the sum of at most positive squares. Therefore,
cannot be expressed as the sum of
positive squares, which is a contradiction. Hence,
for each
.
(b) To find an integer such that
, we need to show that
can be expressed as the sum of
positive squares.
Consider the number . We can express it as the sum of
perfect squares of
and
perfect square of
. Therefore,
.
(c) To prove that there are infinitely many integers such that
, note that for any integer
where
is a non-negative integer, we have
. Since there are infinitely many non-negative integers
, there are infinitely many integers
such that
.
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 |