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

1. 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,

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

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

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,

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

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

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

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