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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2024

Paper 4

For fixed $m,l$, set
$$
D_n=F_{n+l}F_{n+m}-F_nF_{n+m+l}.
$$
At $n=0$, $D_0=F_lF_m$. Applying the Fibonacci recurrence to every term and cancelling gives $D_{n+1}=-D_n$. Induction therefore proves
$$
\boxed{F_{n+l}F_{n+m}-F_nF_{n+m+l}
=(-1)^nF_mF_l}.
$$

Applying this identity to the relevant pairs of indices and using $F_{r+1}=F_r+F_{r-1}$ gives
$$
F_{j+k}^2-F_{j-k}^2=F_{2k}F_{2j}
$$
and
$$
F_{j+k+1}^2+F_{j-k}^2=F_{2k+1}F_{2j+1}.
$$
Thus, for $j\geq k\geq0$,
$$
\boxed{(F_{j+k}-F_{j-k})(F_{j+k}+F_{j-k})=F_{2k}F_{2j}}
$$
and
$$
\boxed{F_{j+k+1}^2+F_{j-k}^2=F_{2k+1}F_{2j+1}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 6E / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 6E / ii / Solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 6E / ii / Solution

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