Question

One leg of a right triangle has length 7 and all sides are whole numbers. Find the lengths of the other two sides.

Answer 1

\(\large \color{black}{\begin{align}

x^2=y^2+7^2 \hspace{.33em}\\~\\
x^2-y^2=49 \hspace{.33em}\\~\\
(x-y)(x+y)&=49\times 1 \hspace{.33em}\\~\\
(x-y)(x+y)&=1\times 49 \hspace{.33em}\\~\\
(x-y)(x+y)&=7\times 7 \hspace{.33em}\\~\\

\end{align}}\)

Answer 2

as they are whole numbers only three types of system of equations are possible

\(\large \color{black }{\begin{align}
(x-y)=49 \hspace{.33em}\\~\\
(x+y)=1\hspace{.33em}\\~\\

\end{align}}\)

or

\(\large \color{black }{\begin{align}
(x-y)=1 \hspace{.33em}\\~\\
(x+y)=49\hspace{.33em}\\~\\

\end{align}}\)

or

\(\large \color{black }{\begin{align}
(x-y)=7 \hspace{.33em}\\~\\
(x+y)=7\hspace{.33em}\\~\\

\end{align}}\)