234. Find a square number which, when increased or decreased by 5, remains a square number.
234. Leonardo's Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-133.jpg?height=554&width=611&top_left_y=480&top_left_x=371)

Fig. 55. Let the required number be \( x^{2} \), then by the condition:

\[
x^{2}+5=u^{2} \text { and } x^{2}-5=v^{2}
\]

from which

\[
\begin{aligned}
u^{2}-v^{2} & =10=(u+v)(u-v)= \\
& =\frac{1440}{144}=\frac{80 \cdot 18}{12^{2}}
\end{aligned}
\]

Assuming

\[
u+v=\frac{80}{12} \text { and } u-v=\frac{18}{12}
\]

we find:

\[
u=\frac{49}{12} \text { and } v=\frac{31}{12}
\]

from which

\[
x=\frac{41}{12}
\]

Verification:

\[
\left(\frac{41}{12}\right)^{2}+5=\left(\frac{49}{12}\right)^{2} ;\left(\frac{41}{12}\right)^{2}-5=\left(\frac{31}{12}\right)^{2}
\]