Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 6"

(Solution)
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{{solution}}
 
{{solution}}
  
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(a) For any odd number <math>2n+1</math>:
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<math>2n+1 = n^2 + 2n +1 - n^2</math>
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<math>2n+1 = (n+1)^2 - n^2</math>
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(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:
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Case 1: 2n is the difference of two even numbers:
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<math>2n = (2m)^2 - (2p)^2</math>
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<math>2n = 4(m^2 - p^2)</math>
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Therefore 2n is a multiple of 4.
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Case 2: 2n is the difference of two odd numbers:
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<math>2n = (2m+1)^2 - (2p+1)^2</math>
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<math>2n = 4m^2 + 4m + 1 - 4p^2 - 4p - 1</math>
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<math>2n = 4(m(m+1) - p(p+1))</math>
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Therefore 2n is a multiple of 4.
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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 14:42, 28 January 2018

Problem

(a) Demonstrate that every odd number $2n+1$ 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 $2n+1$:

$2n+1 = n^2 + 2n +1 - n^2$

$2n+1 = (n+1)^2 - n^2$

(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:

$2n = (2m)^2 - (2p)^2$

$2n = 4(m^2 - p^2)$

Therefore 2n is a multiple of 4.

Case 2: 2n is the difference of two odd numbers:

$2n = (2m+1)^2 - (2p+1)^2$

$2n = 4m^2 + 4m + 1 - 4p^2 - 4p - 1$

$2n = 4(m(m+1) - p(p+1))$

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 (ProblemsAnswer KeyResources)
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
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