Fibonacci squares

$F_{n+1}^2 - F_n^2 = {\color{#3D99F6}\left(F_{n+1}+F_n\right)}{\color{#D61F06}\left(F_{n+1}-F_n\right)} = {\color{#3D99F6}F_{n+2}}{\color{#D61F06}F_{n-1}}$ . Therefore, $F_{10}^2-F_9^2 = F_{11}F_8 = \boxed{89\cdot 21}$ .

If we replace $n$ with $2$ in the formula given by the hint, then we get ${ ({ F }_{ 3 }) }^{ 2 }-{ ({ F }_{ 2 }) }^{ 2 }$ , which can be simplified to ${ (2) }^{ 2 }-{ (1) }^{ 2 }$ , which is equal to $3\cdot 1$ .

Replacing $n$ with $3$ , we get ${ ({ F }_{ 4 }) }^{ 2 }-{ ({ F }_{ 3 }) }^{ 2 }$ , or $5\cdot 1$ .

Replacing it with $4$ , we get ${ ({ F }_{ 5 }) }^{ 2 }-{ ({ F }_{ 4 }) }^{ 2 }$ , or $8\cdot 2$ .

Interestingly, all of the numbers we get are Fibonacci numbers multiplied by other Fibonacci numbers. Using this fact, we can continue the pattern without having to do very many calculations:

$\begin{bmatrix} { ({ F }_{ 3 }) }^{ 2 }-{ ({ F }_{ 2 }) }^{ 2 } & { 2 }^{ 2 }-{ 1 }^{ 2 } & 3\cdot 1 \\ { ({ F }_{ 4 }) }^{ 2 }-{ ({ F }_{ 3 }) }^{ 2 } & { 3 }^{ 2 }-{ 2 }^{ 2 } & 5\cdot 1 \\ { ({ F }_{ 5 }) }^{ 2 }-{ ({ F }_{ 4 }) }^{ 2 } & { 5 }^{ 2 }-{ 3 }^{ 2 } & 8\cdot 2 \\ { ({ F }_{ 6 }) }^{ 2 }-{ ({ F }_{ 5 }) }^{ 2 } & { 8 }^{ 2 }-{ 5 }^{ 2 } & 13\cdot 3 \\ { ({ F }_{ 7 }) }^{ 2 }-{ ({ F }_{ 6 }) }^{ 2 } & { 13 }^{ 2 }-{ 8 }^{ 2 } & 21\cdot 5 \\ { ({ F }_{ 8 }) }^{ 2 }-{ ({ F }_{ 7 }) }^{ 2 } & { 21 }^{ 2 }-{ 13 }^{ 2 } & 34\cdot 8 \\ { ({ F }_{ 9 }) }^{ 2 }-{ ({ F }_{ 8 }) }^{ 2 } & { 34 }^{ 2 }-{ 21 }^{ 2 } & 55\cdot 13 \\ \boxed { { ({ F }_{ 10 }) }^{ 2 }-{ ({ F }_{ 9 }) }^{ 2 } } & 55^{ 2 }-34^{ 2 } & \boxed { 89\cdot 21 } \end{bmatrix}$

$F_{10} = 55, F_9 = 34 \Rightarrow F_{10} ^ 2 - F_9 ^ 2 = 3025 - 1156 = 1869$ .

The last digit is $9$ , so we can just find the last digit is 9.

$55 \times 13 ^ 2 = 55 \times 169 \equiv 5 ( \mod 10 )$

$89 \times 21 \equiv 9 ( \mod 10 )$

$34 ^ 2 \equiv 6 ( \mod 10 )$

$55 \times 13 \equiv 5 ( \mod 10 )$

$34 ^ 2 + 13 = 1156 + 13 \equiv 9 ( \mod 10 )$

$34 ^ { 2 } + 13 ^ { 2 } = 1156 + 169 \equiv 5 ( \mod 10 )$

$55 \times 21 \equiv 5 ( \mod 10 )$

$21 ^ { 2 } = 1 ( \mod 10 )$

So, $89 \times 21$ , or $34 ^ { 2 } + 13$ can be the answer.

But, $89 \times 21 = 1869$ , and $34 ^ { 2 } + 13 = 1169$ .

So $\boxed { 89 \times 21 }$ is the answer.