Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 6"
Mathisfun04 (talk | contribs) (→Solution) |
Mitko pitko (talk | contribs) |
||
| Line 8: | Line 8: | ||
{{solution}} | {{solution}} | ||
| + | (a) For any odd number <math>2n+1</math>: | ||
| + | |||
| + | <math>2n+1 = n^2 + 2n +1 - n^2</math> | ||
| + | |||
| + | <math>2n+1 = (n+1)^2 - n^2</math> | ||
| + | |||
| + | (b) When expressing an even number as a difference of two numbers, these must be of the same parity, i.e. both must be either even or odd. The parity of a number stays the same when squaring it. This gives us two cases: | ||
| + | |||
| + | Case 1: 2n is the difference of two even numbers: | ||
| + | |||
| + | <math>2n = (2m)^2 - (2p)^2</math> | ||
| + | |||
| + | <math>2n = 4(m^2 - p^2)</math> | ||
| + | |||
| + | Therefore 2n is a multiple of 4. | ||
| + | |||
| + | Case 2: 2n is the difference of two odd numbers: | ||
| + | |||
| + | <math>2n = (2m+1)^2 - (2p+1)^2</math> | ||
| + | |||
| + | <math>2n = 4m^2 + 4m + 1 - 4p^2 - 4p - 1</math> | ||
| + | |||
| + | <math>2n = 4(m(m+1) - p(p+1))</math> | ||
| + | |||
| + | Therefore 2n is a multiple of 4. | ||
| + | |||
| + | For both cases we find that only even numbers which are multiples of 4 can be expressed as a difference of two squares. | ||
== See Also == | == See Also == | ||
{{UNCO Math Contest box|n=II|year=2007|num-b=5|num-a=7}} | {{UNCO Math Contest box|n=II|year=2007|num-b=5|num-a=7}} | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 13:42, 28 January 2018
Problem
(a) Demonstrate that every odd number 
can be expressed as a difference of two squares.
(b) Demonstrate which even numbers can be expressed as a difference of two squares.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
(a) For any odd number :
(b) When expressing an even number as a difference of two numbers, these must be of the same parity, i.e. both must be either even or odd. The parity of a number stays the same when squaring it. This gives us two cases:
Case 1: 2n is the difference of two even numbers:
Therefore 2n is a multiple of 4.
Case 2: 2n is the difference of two odd numbers:
Therefore 2n is a multiple of 4.
For both cases we find that only even numbers which are multiples of 4 can be expressed as a difference of two squares.
See Also
| 2007 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
