Question 120479
show that the following statement is {{{incorrect}}}:


“{{{The}}}{{{ sum}}}{{{ of }}}{{{any}}}{{{ two }}}{{{odd }}}{{{numbers}}}{{{ is}}}{{{ divisible}}}{{{ by }}}{{{4}}}."


first recall some of {{{properties}}} relating to {{{odd}}} numbers:

-All odd numbers can be expressed as {{{2n+1}}} where {{{n}}} is a whole number.

-Sum or difference of {{{2 }}}{{{odd}}} numbers is {{{always}}}{{{ even}}}. 

-Sum of {{{odd }}}and {{{even}}} number is {{{always}}}{{{ odd}}}.
	

also recall that: 


-A number {{{is}}}{{{ divisible}}} by {{{4}}}, when the number formed by the last two right hand digit is divisible by {{{4}}}. 

-Or, a number is {{{divisible}}} by {{{4}}}, if its two last digits are {{{zeros}}} or they make a {{{two-digit}}}{{{ number}}}, which is divisible by {{{4}}}.


Any integer {{{n}}} can be put into {{{one}}} of the four cases {{{4q}}}, {{{4q+1}}}, {{{4q+2}}}, and {{{4q+3}}}. 

Since {{{4q}}} and {{{4q+2}}} are {{{even}}}, only the cases {{{4q+1}}} and {{{4q+3}}} need be considered. 


{{{SUPPOSE}}} that {{{x }}}is the {{{EVEN}}} number. Then {{{x}}} has a factor of {{{2}}} and {{{x^2}}} has a factor of {{{4 }}}; that is, {{{x^2}}}  is divisible by {{{4 }}}.

	
If we choose {{{two}}}{{{ odd}}} numbers, let’s say {{{1}}} and {{{3}}}, and {{{add }}}their squares ({{{1^2 + 3^2}}} in this case) we will find that the {{{sum}}} is {{{not}}}{{{ divisible}}} by {{{4}}}. 

{{{1^2 + 3^2 = 1 + 9 = 10}}}….{{{10/4= 2.5}}}…… => ..{{{not}}}{{{ divisible}}} by {{{4}}}. 


If{{{ a}}} and {{{b}}} are {{{both }}}{{{odd}}} they have the form {{{a = 2 x + 1}}} and {{{b =  y + 1}}}. 


Then {{{a^2 + b^2 = (2 * x + 1)2 + (2 y + 1)2 = 4 ื (x2 + y2 + x + y) + 2}}}, which has a {{{remainder}}} of {{{2}}} when divided by {{{4}}} and so can not be equal to {{{x^2}}}, which is exactly divisible by {{{4}}}. 


Therefore the {{{ASSUMPTION}}}that {{{x}}} is {{{EVEN}}} is {{{INCORRECT}}}. 

Since {{{we}}}{{{ showed}}} that {{{one }}}of the terms {{{MUST}}}{{{ be}}}{{{ even}}}, one of {{{a}}} and {{{b}}} {{{MUST}}}{{{ be}}}{{{ even}}}, and {{{the }}}{{{other}}}{{{ odd}}}.