Flos/Problems/Problem 1

Problem from Flos by Leonardo Fibonacci

Find a rational number such that $5$ added to, or subtracted from, its square, is also a square.

Solution

$\dfrac {41} {12}$

Proof

We need to find $3$ squares in arithmetic sequence whose common difference is of the form $5 p^2$.

That is:

$\tuple {x^2 - 5 p^2, x^2, x^2 + 5 p^2}$

We can find:

\(\ds 41^2 - 5 \times 12^2\)

\(=\)

\(\ds 1681 - 720\)

\(\ds \)

\(=\)

\(\ds 961\)

\(\ds \)

\(=\)

\(\ds 31^2\)

and:

\(\ds 41^2 + 5 \times 12^2\)

\(=\)

\(\ds 1681 + 720\)

\(\ds \)

\(=\)

\(\ds 2401\)

\(\ds \)

\(=\)

\(\ds 49^2\)

Hence:

\(\ds \paren {\dfrac {31} {12} }^2\)

\(=\)

\(\ds \paren {\dfrac {41} {12} }^2 - 5\)

\(\ds \paren {\dfrac {49} {12} }^2\)

\(=\)

\(\ds \paren {\dfrac {41} {12} }^2 + 5\)

$\blacksquare$

Sources

• 1225: Leonardo Fibonacci: Flos
• 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Liber Abaci: $86$