Prove that the sum of the squares of three odd integers

- 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,

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

expanding

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

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

not sure where to go from here.

Re: Prove that the sum of the squares of three odd integers

Quote:

Originally Posted by

**Tweety** - 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,

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

expanding

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

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

not sure where to go from here.

you're so close...

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

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

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