Sequence proof if it exists For every integer r>=3 there exists a sequence $a_{1,\space }a_{2,}.....,a_r$ of nonzero integers with the property that
$a_1^2+a_2^2+....+a_{r-1}^2=a_r^2$
I tried to prove this with proof by induction (as to whether it was false or not) I'm not sure if that was the correct method. any help with solving this would be greatly appreciated
 A: Well, let's see.
$$
3^2+4^2=5^2
$$
so if there exist
$$
a_1^2+a_2^2+...+a_r^2=a_{r+1}^2
$$
then
$$
\left(\frac{a_1}{a_{r+1}}\right)^2+...+\left(\frac{a_r}{a_{r+1}}\right)^2=1
$$
so
$$
\left(\left(\frac{a_1}{a_{r+1}}\right)^2+...+\left(\frac{a_r}{a_{r+1}}\right)^2\right)\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=1
$$
then we can just multiply both sides by $(5a_{r+1})^2$ and get the $r+1$ step. 
A: $$\begin{cases}3^2+4^2=5^2\\a_1^2+a_2^2+\cdots+a_r^2=a_{r+1}^2\end{cases}\iff (3a_1)^2+(3a_2)^2+\cdots+(3a_r)^2=3^2a_{r+1}^2=(5^2-4^2)a_{r+1}^2$$
$$\iff(3a_1)^2+(3a_2)^2+\cdots+(3a_r)^2+(4a_{r+1})^2=(5a_{r+1})^2$$
Thus we've proved this using induction. $\ \ \ \square$
This is a simplified version of Hans Spielgarten's answer (which unnecessarily uses division).
A: Hint 
$$2k+1=(k+1)^2-k^2 \\
4k=(k+1)^2-(k-1)^2$$
Use this to show that $a_r^2$ is the difference of two perfect squares. This proves immediately the inductive step.
A: Let us start with the good old $3^2+4^2=5^2$, that is, $a_1=3$, $a_2=4$, and $a_3=5$, which gives a sequence of length $3$,  and produce a sequence of length $4$. 
The idea is to multiply the first two terms of our previous sequence  through by $3$, and add terms $4\cdot 5$ and $5\cdot 5$.  So for $r=4$ our sequence is $3\cdot 3$, $3\cdot 4$, $4\cdot 5$, and $5\cdot 5$. 
For a sequence of length $5$, multiply the first three terms of the previous sequence by $3$, and add $4\cdot 5\cdot 5$ and $5\cdot 5\cdot 5$. 
Continue.  
A: Hint:
Divide by $a_{r}^2$ and the equivalent equation is
$$\left(\frac{a_1}{a_{r}}\right)^2+ \cdots +\left(\frac{a_{r-1}}{a_{r}}\right)^2 =1$$
a rational point on the unit sphere $S^{r-2}=\{\sum_{i=1}^{r-1} x_i^2=1\}$. Use the stereographic projection from the South pole that gives a bijection between the rational points on  $S^{r-2}\backslash{\text{South Pole}}$ and $\mathbb{Q}^{r-2}$
$$\mathbb{Q}^{r-2} \ni t \mapsto \left( \frac{2 t}{1+|\,t|^2}, \frac{1-|t|^2}{1+|t|^2}\right)=(x', x_{r-1})\in S^{r-2}\cap \mathbb{Q}^{r-1}$$
with inverse 
$$  t = \frac{x'}{1+x_{r-1}}$$
For example, take $r=5$. 
$$\mathbb{Q}^3\ni (1,2,5) \mapsto \left(\frac{2}{31},\frac{4}{31},\frac{10}{31}, -\frac{29}{31}\right)$$
and so
$$2^2 + 4^2 + 10^2 + 29^2 = 31^2$$
This also makes for a quick solution: start with any $r-2$ integers $(a_1, \ldots, a_{r-2})$. Denote the vector $(a_1, \ldots , a_{r-2})$ by $t$. Take any integer $b$. Then $(2b t, b^2-|t|^2, b^2+|t|^2)$ is a solution to our problem:
$$(2b a_1)^2 + \cdots + (2ba_{r-2})^2 + (b^2 - (a_1^2 + \cdots + a_{r-2}^2))^2 = (b^2 + (a_1^2 + \cdots + a_{r-2}^2))^2$$
