TITLE: Let $p_1, p_2,\dots,p_n$ be polynomials of $k$ variables $x_1,\dots,x_k$ and $p_1^2 + \dots +p_n^2=x_1^2 + \dots + x_k^2$ Prove that $n \geq k$.
QUESTION [25 upvotes]: Let $p_1, p_2,...,p_n$ be real polynomials of $k$ variables $x_1,...,x_k$ and assume that 
$$p_1^2 + \dots +p_n^2=x_1^2 + \dots + x_k^2$$ 
Prove that $n \geq k$. 
Out of so many questions that I have attempted, this is the first question that I have absolutely no idea how to start (I am ashamed of myself). Can anyone give some hint?

REPLY [3 votes]: Having written everything out, I now realize that this merely a paraphrasing of leonbloy's answer, using explicit indexed sums and filling out the details, but figure I might as well post it at this point. 
As has already been pointed out, each polynomial must be strictly linear. That is, for each $q$, we may write $$p_q(\mathbf{x})=\sum_{i=1}^k \alpha_q^ix_i,$$ so that $\alpha_q^i$ is the scalar of $x_i$ in the $q^{\text{th}}$ polynomial. But then $$\sum_{q=1}^n(p_q(\textbf{x}))^2=\left(\sum_{i=1}^k \left[\sum_{q=1}^n (\alpha_q^i)^2\right](x_i)^2\right)+\left(\sum_{i=1}^k \sum_{j=1}^k \left[\sum_{q=1}^n 2\alpha_q^i\alpha_q^j\right]x_ix_j\right).$$
Thus, if these polynomials $p_1,p_2,\ldots,p_n$ satisfy the property that $\sum_{q=1}^n(p_q(\textbf{x}))^2=\sum_{i=1}^k (x_i)^2$, then necessarily
$$\sum_{q=1}^n (\alpha_q^i)^2=1$$
for all $i$, while $$\sum_{q=1}^n 2\alpha_q^i\alpha_q^j=0$$ for all $i\neq j$. Thinking of these $\alpha_i^q$ as indexed vectors in $n-$dimensional euclidean space, we see that this boils down to the statement: "you can't have $q$ orthogonal unit vectors in $\mathbb{R}^n$ unless $n\geq q$", which follows from the basis theorem of linear algebra. I hope that this makes sense. Please let me know if anything here is unclear at all. Great question!