A proof question using mathematical Induction 
I'm having difficulty as to understanding what the questions is expecting us to prove. I understand how the mathematical induction works and how to go about doing the same. It would help if someone could in simpler words explain what the question means.
 A: The base of the induction is obvious.
Let $a^2=a_1^2+...+a_n^2$.
Hence, $(2a)^2=(2a_1)^2+...+(2a_n)^2$.
Thus, there are naturals $p$ and $q$  for which $2a=2pq$ and $p>q$.
Thus,
$$(p^2+q^2)^2=(2pq)^2+(p^2-q^2)^2=(2a_1)^2+...+(2a_n)^2+(p^2-q^2)^2$$
and we are done!
A: The question states that, when choosing any natural number $n\ge2$, you can find $n+1$ integers $a$, $a_1$, ..., $a_n$ such that 
$$a^2 = a_1^2 + \dots + a_n^2$$
For instance, for $n=2$:
$$5^2 = 4^2 + 3^2.$$
That would be the first step for the induction process. Now assume that the statement is satisfied for any number up to $n-1$ and prove it for $n$...
A: You must prove that you can find positive integers $a$ and $a_i$ such that
$$a^2=a_1^2+a_2^2$$
and (not necessarily the same)
$$a^2=a_1^2+a_2^2+a_3^2$$
and also
$$a^2=a_1^2+a_2^2+a_3^2+a_4^2$$
and so on, ad libitum.

For example,
$$5^2=3^2+4^2,\\13^2=3^2+4^2+12^2,\\2^2=1^2+1^2+1^2+1^2,\\\cdots$$
A: You need to prove that there is some square that can be written as the sum of two squares, some square that can be written as the sum of three squares, some square that can be written as the sum of four squares and so on and so on...
