fixes to 9.1.21(c), 4.1.21, 4.5.17

Author: spamegg1

src/Epp.tex

@@ -10490,7 +10490,7 @@ \subsubsection{Exercise 21}
 
 $20 = 4^2 + 2^2$
 
-$22 = 2^2 + 2^2 + 3^2$
+$22 = 2^2 + 3^2 + 3^2$
 
 $24 = 4^2 + 2^2 + 2^2$
 \end{proof}
@@ -12870,7 +12870,7 @@ \subsubsection{Exercise 17}
 
 By definition of even, $n = 2k + 1$ for some integer $k$.
 
-Then $n^2-n + 3 = (2k+1)^2 - 2(k+1) + 3 = 4k^2 + 4k + 4 - 2k - 2 + 3 = 4k^2 + 2k + 5 = 2(2k^2 + k + 2) + 1$.
+Then $n^2-n + 3 = (2k+1)^2 - (2k+1) + 3 = 4k^2 + 4k + 4 - 2k - 2 + 3 = 4k^2 + 2k + 5 = 2(2k^2 + k + 2) + 1$.
 
 Let $t = 2k^2 + k + 2$, which is an integer because it is a sum and product of integers. So $n^2 - n + 3 = 2t+1$ where $t$ is an integer. So by definition of odd, $n^2 - n + 3$ is odd.
 \end{proof}
@@ -40182,7 +40182,7 @@ \subsubsection{Exercise 21}
 \begin{proof}
 Of the integers from 10 through 99 that are multiples of 4, the smallest is 12 \((= 4 \cdot 3)\) and the largest is 96
 \((= 4 \cdot 24)\). Thus there are \(24 - 3 + 1 = 22\) two-digit integers that are multiples of 4. Hence the
-probability that a randomly chosen two-digit integer is a multiple of 4 is \(22/90 \approx 36.6\%\).
+probability that a randomly chosen two-digit integer is a multiple of 4 is \(22/90 \approx 24.4\%\).
 \end{proof}
 
 \subsubsection{Exercise 22}