I'm trying to solve an induction proof exercise but this time I can't even understand how to proceed. I must prove that for every given $n\in \mathbb{N}$ with $n\geq2$ there exist $a,a_1,a_2,...,a_n$ in $\Bbb N\setminus\{0\}$ so that:
$$a^2 = a_1^2 + a_2^2 +\ldots+ a_n^2$$
I can't even think about the base case.
Can someone help me solving this problem please?
On
HINT: The base case is $n=2$: you must find positive integers $a,a_1$, and $a_2$ such that $a^2=a_1^2+a_2^2$. All you need is one example of this; think about familiar right triangles.
For the induction step step you’ll assume that you’ve found positive integers $a,a_1,\ldots,a_n$ such that
$$a^2=a_1^2+a_2^2+\ldots+a_n^2\;,\tag{1}$$
and you’ll try to use that to show that there are positive integers $b,b_1,\ldots,b_{n+1}$ such that
$$b^2=b_1^2+b_2^2+\ldots+b_{n+1}^2\;.$$
Note that if $k$ is any positive integer, and you multiply $(1)$ by $k^2$, you get another solution for $n$:
$$(ka)^2=(ka_1)^2+(ka_2)^2+\ldots+(ka_n)^2\;.$$
You can use this idea to make $(ka_n)^2$ a number that can be written as the sum of two squares, then set $b=ka$, $b_1=ka_1$, and so on up to $b_{n-1}=ka_{n-1}$.
Induction Step: Suppose we know that there exist positive integers $a$, $a_1,a_2,\dots a_k$ such that $a^2=a_1^2+a_2^2+\cdots+a_k^2$.
For $i=1$ to $k$, let $b_i=3a_i$, and let $b_{k+1}=4a$. Let $b=5a$. Then $$b_1^2+b_2^2+\cdots+b_k^2+b_{k+1}^2=9a^2+16a^2=(5a)^2=b^2.$$