# To prove: The given statement, " If k is any odd integer and m is any even integer, then k^{2} + m^{2} is odd".

Transformation properties
To prove:
The given statement, " If k is any odd integer and m is any even integer, then $$\displaystyle{k}^{{{2}}}\ +\ {m}^{{{2}}}$$ is odd".

2020-11-18
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.