If $D$ is an $\mathbb{R}$-algebra that is also a division ring and $\dim_{\mathbb{R}}D=n<\infty$, then for every $d\in D$ there exists $\lambda\in\mathbb{R}$ such that $d^2+\lambda d\in\mathbb R$.
Is the statement equivalent to $d^2+\lambda{}d=0$? Otherwise I cannot make of it being a real number.
I tried setting $d=\sum_{i=1}^n\lambda_id_i$ but it doesn't give me anything, how can I show it?
The statement is bizzare stated that way since $d^2+\lambda d\in D$. I assume it is meant that there exist $a,b\in\mathbb{R}: d^2+ad+b\cdot1=0,\forall d\in D$.
For every $d\in D$ consider the $n+1$ elements $1,d,...,d^n$. By linear dependence there exist: $a_1,...,a_n\in\mathbb{R}: a_0\cdot1+a_1d+\dots+a_nd^n=0 \implies d^n+\dots +b_0\cdot1=0$ where $b_i=a_i/a_n$ (if $a_n=0$ take the largest $a_i\ne{}0$).
Consider the monic polynomial $p(x)=x^n+b_{n-1}x^{n-1}+\dots+b_0$. In $\mathbb{R}$ it can be factored as $p(x)=p_1(x)p_2(x)\cdots p_k(x)$ where $\deg(p_i)\leq{}2$.
Since $p(d)=0$ we have $p_j(d)=0$ for some $j$, implying $d=r\cdot1,r\in\mathbb{R}$ or $d^2+ad+b\cdot1=0,a,b\in\mathbb{R}$.