0
$\begingroup$

My question is the following: Show that for each integer $k \ge 5$, there is an integer $N(k)$ such that every integer $n \ge N(k)$ can be written as a sum of $k$ nonzero squares.

What is the process to prove this? I'm able to do this for the first few values of $k$, but I'm not able to find a version of induction that works for this.

$\endgroup$
1
$\begingroup$

You state that you've already proved this for "the first few values of $k$", so only the inductive step remains.

1 is a perfect square. Hence, if you can write $n$ as a sum of $k$ nonzero squares, then you can write $n+1$ as a sum of $k+1$ nonzero squares, by letting one of those squares equal 1 and thus reducing to a previously solved problem.

It then follows that $N(k+1) \le N(k)+1$.

$\endgroup$
  • $\begingroup$ You have to prove the base case $k=5$ to make this induction work. $\endgroup$ – TonyK Nov 8 '18 at 2:51
  • $\begingroup$ @TonyK Indeed you do, but OP says "I'm able to do this for the first few values of k" so presumably that's taken care of. I will edit my post to clarify this. $\endgroup$ – Geoffrey Brent Nov 8 '18 at 2:52
  • $\begingroup$ Unlikely the OP has a proof for 5 nonzero squares. See my brief answer $\endgroup$ – Will Jagy Nov 8 '18 at 3:33
0
$\begingroup$

In the last few pages of The Sensual Quadratic Form by J. H. Conway, it is proved that the numbers that are not the sum of four nonzero integer squares are $$ 1,3,5,9,11,17,29,41, \; 2 \cdot 4^m \; , \; 6 \cdot 4^m \; , \; 14 \cdot 4^m $$

ADDED: with four nonzero squares we get all odd numbers 43 or larger. Add a single 1 and we get all even numbers 44 or larger. Alternatively, add a single 4, get all odd numbers 47 or above with five nonzero squares. Together, we can write all numbers 47 or larger with five nonzero squares.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.