Answer
We wish to prove that the square of any odd integer has the form $8m+1$ for some integer $m$.
Let $n$ be an odd integer. By the definition of odd, $n=2k+1$ for some integer $k$. Thus, $n^{2}=(2k+1)^{2}=4k^{2}+4k+1$$=4(k^{2}+k)+1=4[k(k+1)]+1$. But $k(k+1)$ is a product of two consecutive integers, so by the result of the previous exercise, it must be even. By the definition of even, we therefore have $k(k+1)=2m$ for some integer $m$. By substitution, we get $n^{2}=4[k(k+1)]+1=4(2m)+1=8m+1$, which is precisely in the desired form. Since $n$ was arbitrarily chosen, we conclude that the square of any odd integer may be written in the form $8m+1$ for some integer $m$.
Work Step by Step
The "blind alley" referred to in the problem was that, until the previous exercise, we did not know whether $k^{2}+k$ was even or odd. See the paragraph beginning at the bottom of page 185 and continuing onto page 186.
