# Prove that $10^{(n+k)} - 10 ^ n$can be written as sum of $18k$ perfect (nonzero) squares

Prove that $$10^{n+k} - 10^{k}$$ can be written as a sum of $$18k$$ perfect (non-zero) squares; for every $$n \ge 1,\: k \ge 1$$.

Is there a way to prove this without using induction, using elementary school techniques? (induction solution below)

Thanks!

If you consider a telescoping sum so obviously correct it's not "induction",$$10^{n+k}-10^n=\sum_{j=0}^{k-1}(10^{n+j+1}-10^{n+j})=\sum_{j=0}^{k-1}9\times10^{n+j},$$so you just need to write $$10^{n+j}$$ as the sum of two squares. If $$n+j$$ is odd, multiply $$10=1^2+3^2$$ by $$10^{n+j-1}=(10^{(n+j-1)^2})^2$$; if $$n+j$$ is even it's $$\ge2$$, so multiply $$100=6^2+8^2$$ by $$10^{n+j-2}=(10^{(n+j)/2-1})^2$$.

I got one solution by induction, for n odd and n even;

E.g. if n is odd, $$10^n = 10^{n-1} \times 10 \times (10^{k} - 1)$$;

Showing by induction that $$10(10^k - 1) = p_{1}^{2} + ... p_{18k}^{2}$$:

P(1) : $$90 = 9 \times 1 + 9 \times 9$$ --> 18 perfect squares

P(2) : $$990 = 9 \times 1 + 9 \times 9 + 9 \times 6^{2} + 9\times 8^{2}$$--> 36 perfect squares Assuming P(k) true, we prove P(k+2), by replacing 10^k with the sum of squares from P(k) + 1, and rearranging to obtain 18k +36 squares.

And, similarly, for n even (showing that $$100 * (10^k - 1)$$ can be written as a sum of 18k squares.