Let $R$ be a ring (not necessarily commutative). Let $u\in R^\times$, (so $u$ is a unit), and let $a\in R$ be a nilpotent element. Show that if $ua=au$ then $u+a\in R^\times$.  I have that if $a$ is nilpotent, then $1-a$ is a unit. Then, I know that this means $\exists b\in R$ such that $ (1-a)b=1$. Additionally I have that $\exists c\in R$ such that $uc=1$. Then I was thinking maybe I could do some kind of algebraic manipulation like $uc=(1-a)b$ but that seems to only lead to dead ends.  Any hints or pointers in the right direction appreciated.
If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.   I've been working on this problem for an hour that I tried to construct an element $x \in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?
This is question 2.9 #9 from Topics in Algebra by Herstein:     If $o(G)$ is $pq$ where $p$ and $q$ are distinct prime numbers and if $G$ has a normal subgroup of order $p$ and a normal subgroup of order $q$, prove that $G$ is cyclic.   Let $N$ denote the normal subgroup of order $p$ and $M$ denote the normal subgroup of order $q$. Here are a couple things I noted while exploring the problem:   $N$ and $M$ are cyclic, since they have prime order. $G = NM$ [1].   Any hints where to look next?    [1] Since $N,M$ are normal, $NM \le G$. Then $o(NM) \mid o(G)$, so $o(NM)$ is either $1$, $p$, $q$, or $pq$. The first three don't work because they force one of $N$ or $M$ to be $(e)$, so we have $o(NM) = pq$. Therefore, $NM = G$. We know that $12^2 = 144$ and that $38^2 = 1444$. Are there any other perfect squares in the form of $\frac{13}{9} (10^n - 1) + 1$ (i.e. $1$ followed by $n$ $4$'s), and how would we prove it? 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For instance, say you have:  $$c_0 = 20$$ $$c_1 = 30$$ $$c_n = 3 c_{n-1} - 2 c_{n-2}$$  We can write the characteristic equation as:  $$3x^{n-1} - 2x^{n-2} = x^n$$  Solving this with $n=2$, we get $x = 1$ or $x = 2$. This lets us write the relation $c_n = \alpha_1 1^n + \alpha_2 2^n$, and we can solve for $\alpha_1$ and $\alpha_2$ with the initial states $c_0$ and $c_1$.  However, this depends on the fact that $3x^{n-1} - 2x^{n-2} = x^n$ has two roots.  Now, I'm stuck on another problem where the characteristic equation has fewer roots than terms.  Say I have this recurrence relation instead:  $$a_0 = 0$$ $$a_1 = 2$$ $$a_2 = −1$$ $$a_n = 9a_{n-1} - 15a_{n-2} - 25a_{n-3}$$  The characteristic equation would be:  $$9x^{n-1} - 15x^{n-2} - 25x^{n-3} = x^n$$  However, solving with $n=3$, we only get two roots: $x=-1$ or $x=5$. There are not enough roots to write a relation in the form of $a_n = \alpha_1 r_1^n + \alpha_2r_2^n + \alpha_3r_3^n$. How do I proceed? What is the meaning of this "let there be light" joke? Windows 10 April 2018 Update added an extra Language and I cannot remove it I have just updated to Windows 10 April 2018 Update (1803 / 17134.1)  I had my laptop configured completely as UK with only a UK keyboard layout  Since the update, a US keyboard has been added so I now see the switch icon in my task bar    It appears that all the keyboard language settings are now migrated to the new Settings app and no longer in the old Control Panel    However, as can be seen from the screenshot above, the Remove button under the newly added US layout is greyed out (as is the UK one)  How can I remove the US layout?  UPDATE 1: I just updated my desktop and it hasn't added a US keyboard!  UPDATE 2: I added a third language (French) and was able to remove it, the button was enabled. The US remove button remained disabled both after adding French and after removing it.  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From here it's easy to see that $ \rho(D_A+D_B) \le \rho(D_A) + \rho(D_B) $, since $ |\lambda_i+\mu_i|\le|\lambda_i|+|\mu_i| \ \forall i$. But I don't see how to include $\rho(A+B)$ in this.  And the last part of the exercise is proving that if $ \{X_i\}$ is a basis of eigenvectors, which is common to A and B, then $ \rho(A+B) \le \rho(A) + \rho(B) $.  Here's my attempt, but I would like to know if it's correct:  Let $\lambda_i$ and $\mu_i$ be eigenvalues of A and B associated to the vector $X_i$, then $AX_i=\lambda_i X_i$ and $BX_i = \mu_iX_i$, and if we sum both equalities, we have $AX_i + BX_i = \lambda_iX_i + \mu_iXi$ so $(A+B)X_i=(\lambda_i+\mu_i)X_i$ so $(\lambda_i+\mu_i)$ is an eigenvalor of $(A+B)$ associated to the eigenvector $X_i$. 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