Sums of Squares problem.

Printable View

April 6th 2008, 08:36 PM
Proof_of_life

Sums of Squares problem.

Let $n = 17, 51, 37, 407, 629, 40885.$

(a) Decide whether $n \in S_2$ .

(b) If $n \in S_2$ find $a,b \in \mathbb{Z}$ with $n=a^2+b^2.$

(c) If $n \notin S_2$ find $a,b,c,d \in \mathbb{Z}$ with $n=a^2+b^2+c^2+d^2.$

Thank you all for checking this out, and any ideas/suggestions are very welcome. Thanks!
April 6th 2008, 08:38 PM
Jhevon

Quote:

Originally Posted by Proof_of_life

Let $n = 17, 51, 37, 407, 629, 40885.$

(a) Decide whether $n \in S_2$ .

(b) If $n \in S_2$ find $a,b \in \mathbb{Z}$ with $n=a^2+b^2.$

(c) If $n \notin S_2$ find $a,b,c,d \in \mathbb{Z}$ with $n=a^2+b^2+c^2+d^2.$

Thank you all for checking this out, and any ideas/suggestions are very welcome. Thanks!

$S_2$ is the set of all numbers that are the sum of the squares of two integers?
April 6th 2008, 09:04 PM
Proof_of_life

$S_2$ is the set of all sums of 2 squares.

Definition:
For each integer $k$ , greater than or equal to 1, let $S_k$ = { $n$ | $n$ = $(x_1)^2$ + . . . + $(x_k)^2$ for some $x_1$ ,..., $x_k \in \mathbb{Z}$ },the set of all sums of $k$ squares.

Sorry, and thanks!
April 6th 2008, 11:33 PM
Opalg

Hint (just to get you started): a square has to be congruent to 0 or 1 mod 4.
April 7th 2008, 01:14 AM
Proof_of_life

So, let me say for the first one that n=17. Since 17 is congruent to 1 mod 4, that would satisfy part (a)? Since part (a) is satisfied, I would take part (b)'s approach and ignore part (c). The answer to part (b) would be $4^2 + 1^2 = 17$ , where a=4 and b=1.

I think I will get confused when n does not belong to $S_2$ , and I will not know how to approach part (c).
April 7th 2008, 01:22 AM
Proof_of_life

Sum of squares

I should have just googled this earlier! Thanks to the last replier, it got me on the right track.

All times are GMT -8. The time now is 06:32 AM.