Difference between revisions of "1997 JBMO Problems/Problem 5"
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== Problem == | == Problem == | ||
| + | |||
| + | Let <math>n_1</math>, <math>n_2</math>, <math>\ldots</math>, <math>n_{1998}</math> be positive integers such that <cmath> n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2. </cmath> Show that at least two of the numbers are even. | ||
== Solution == | == Solution == | ||
| − | == See | + | In order to prove that at least two numbers are even, we need to prove that it is impossible to have exactly one number even and that it is impossible to have exactly no numbers even. |
| + | |||
| + | <br> | ||
| + | '''Lemma 1: Impossible to have exactly 1 number even'''<br> | ||
| + | Assume one of the numbers is even. If <math>n_{1998}</math> is even, then the left side is congruent to <math>1</math> modulo <math>2</math> while the right side is congruent to <math>0</math> modulo <math>2,</math> which can not happen. If one of <math>n_1, n_2, \cdots n_{1997}</math> is even, then the left side is congruent to <math>0</math> modulo <math>2</math> while the right side is congruent to <math>1</math> modulo <math>2,</math> which can not happen. Thus, it is impossible for exactly one of the numbers to be even. | ||
| + | |||
| + | <br> | ||
| + | '''Lemma 2: Impossible to have all numbers odd'''<br> | ||
| + | Assume all numbers are odd (or that there are no even numbers). If <math>n_i</math> is odd for <math>1 \le i \le 1998,</math> then <math>n_i^2 \equiv 1 \pmod{8}.</math> That means the left side is congruent to <math>5</math> modulo <math>8</math> while the right side is congruent to <math>1</math> modulo <math>8.</math> Since this can not happen, it is impossible for none of the numbers to be even. | ||
| + | |||
| + | <br> | ||
| + | Since it’s impossible to have exactly one or zero even numbers, at least two of the integers must be even. | ||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{JBMO box|year=1997|num-b=4|after=Last Problem}} | ||
| − | + | [[Category:Olympiad Number Theory Problems]] | |
Latest revision as of 20:12, 7 August 2018
Problem
Let 
, 
, 
, 
be positive integers such that ![\[n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2.\]](http://latex.artofproblemsolving.com/2/7/9/27994bc5861679d68c68e81da1b333f9770e3752.png)
Show that at least two of the numbers are even.
Solution
In order to prove that at least two numbers are even, we need to prove that it is impossible to have exactly one number even and that it is impossible to have exactly no numbers even.
Lemma 1: Impossible to have exactly 1 number even
Assume one of the numbers is even. If is even, then the left side is congruent to 
modulo 
while the right side is congruent to 
modulo 
which can not happen. If one of 
is even, then the left side is congruent to modulo while the right side is congruent to modulo which can not happen. Thus, it is impossible for exactly one of the numbers to be even.
Lemma 2: Impossible to have all numbers odd
Assume all numbers are odd (or that there are no even numbers). If 
is odd for 
then 
That means the left side is congruent to 
modulo 
while the right side is congruent to modulo 
Since this can not happen, it is impossible for none of the numbers to be even.
Since it’s impossible to have exactly one or zero even numbers, at least two of the integers must be even.
See Also
| 1997 JBMO (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 • 5 | ||
| All JBMO Problems and Solutions | ||
