# Prove that the sum of the squares of three odd integers

• Jan 27th 2014, 05:51 AM
Tweety
Prove that the sum of the squares of three odd integers
1. Prove that the sum of the squares of three odd integers is always 1 less than a multipleof 4.

i am not sure how exactly to prove this, but

i start with let a ,b ,c be odd integers then these can be written as,

$(2a+1)^{2} + (2b+1)^2 + (2c+1)^2$ and want show this expression equals 4k-1

expanding

$4a^{2} +4a + 4b^{2} +4b + 4c^{2} +4c +3$

$4 ( a^{2} +a + b^{2} +b +c^{2} +c ) +3$

not sure where to go from here.
• Jan 27th 2014, 05:59 AM
romsek
Re: Prove that the sum of the squares of three odd integers
Quote:

Originally Posted by Tweety
1. Prove that the sum of the squares of three odd integers is always 1 less than a multipleof 4.

i am not sure how exactly to prove this, but

i start with let a ,b ,c be odd integers then these can be written as,

$(2a+1)^{2} + (2b+1)^2 + (2c+1)^2$ and want show this expression equals 4k-1

expanding

$4a^{2} +4a + 4b^{2} +4b + 4c^{2} +4c +3$

$4 ( a^{2} +a + b^{2} +b +c^{2} +c ) +3$

not sure where to go from here.

you're so close...

$4 ( a^{2} +a + b^{2} +b +c^{2} +c ) +3 =$

$4 ( a^{2} +a + b^{2} +b +c^{2} +c ) + 4-1 =$

$4 ( a^{2} +a + b^{2} +b +c^{2} +c +1)-1$