Proof:

The proof is given as,

\(\displaystyle\text{Let}\ {k}={2}{p}\ +\ {1}\ \text{and}\ {m}={2}{q},\ \text{where}\ {p}\ \text{and}\ {q}\ \text{are integers}.\)

\(\displaystyle{k}^{{{2}}}\ +\ {m}^{{{2}}}={\left({2}{p}\ +\ {1}\right)}^{{{2}}}\ +\ {\left({2}{q}\right)}^{{{2}}}\)

\(\displaystyle={4}{p}^{{{2}}}\ +\ {4}{p}\ +\ {1}\ +\ {4}{q}^{{{2}}}\)

\(\displaystyle={2}\ {\left({2}{p}^{{{2}}}\ +\ {2}{p}\right)}\ +\ {1}\ +\ {2}\ {\left({2}{q}^{{{2}}}\right)}\)

\(\displaystyle={2}{l}\ +\ {1}\ +\ {2}{m}\)

\(\displaystyle\text{where,}\ {l}={2}{p}^{{{2}}}\ +\ {2}{p},\ {m}={2}{q}^{{{2}}}\ \text{are integers.}\)

\(\displaystyle{k}^{{{2}}}\ +\ {m}^{{{2}}}={2}\ {\left({l}\ +\ {m}\right)}\ +\ {1}\)

\(\displaystyle{k}^{{{2}}}\ +\ {m}^{{{2}}}={2}{n}\ +\ {1},\ \text{which is odd by definition.}\)

\(\displaystyle\text{where,}\ {n}={l}\ +\ {m}\ \text{is an integer.}\)

\(\displaystyle\text{Thus,}\ {k}^{{{2}}}\ +\ {m}^{{{2}}}\ \text{is odd.}\)

Conclusion:

The statement, " If k is any odd integer and m is any even integer, then \(\displaystyle{k}^{{{2}}}\ +\ {m}^{{{2}}}\) is odd" is proved.

The proof is given as,

\(\displaystyle\text{Let}\ {k}={2}{p}\ +\ {1}\ \text{and}\ {m}={2}{q},\ \text{where}\ {p}\ \text{and}\ {q}\ \text{are integers}.\)

\(\displaystyle{k}^{{{2}}}\ +\ {m}^{{{2}}}={\left({2}{p}\ +\ {1}\right)}^{{{2}}}\ +\ {\left({2}{q}\right)}^{{{2}}}\)

\(\displaystyle={4}{p}^{{{2}}}\ +\ {4}{p}\ +\ {1}\ +\ {4}{q}^{{{2}}}\)

\(\displaystyle={2}\ {\left({2}{p}^{{{2}}}\ +\ {2}{p}\right)}\ +\ {1}\ +\ {2}\ {\left({2}{q}^{{{2}}}\right)}\)

\(\displaystyle={2}{l}\ +\ {1}\ +\ {2}{m}\)

\(\displaystyle\text{where,}\ {l}={2}{p}^{{{2}}}\ +\ {2}{p},\ {m}={2}{q}^{{{2}}}\ \text{are integers.}\)

\(\displaystyle{k}^{{{2}}}\ +\ {m}^{{{2}}}={2}\ {\left({l}\ +\ {m}\right)}\ +\ {1}\)

\(\displaystyle{k}^{{{2}}}\ +\ {m}^{{{2}}}={2}{n}\ +\ {1},\ \text{which is odd by definition.}\)

\(\displaystyle\text{where,}\ {n}={l}\ +\ {m}\ \text{is an integer.}\)

\(\displaystyle\text{Thus,}\ {k}^{{{2}}}\ +\ {m}^{{{2}}}\ \text{is odd.}\)

Conclusion:

The statement, " If k is any odd integer and m is any even integer, then \(\displaystyle{k}^{{{2}}}\ +\ {m}^{{{2}}}\) is odd" is proved.