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The question is asking me to prove that $F_{1001}$ is congruence to $1 \bmod 4$. I knew that if $n$ is odd then all prime odd divisor $p$ of $F_n$ satisfy $p \equiv 1 \pmod{4}$. However, my question is how to find the odd prime divisor? How do we factor out such a big fibonacci number?

Arturo Magidin
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Algorisum
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2 Answers2

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Factoring big numbers that don't have an immediate product representation is pretty much impossible, without the correspondingly big computers to churn out the numbers. However, we can prove that $F_{1001}\equiv1\pmod{4}$ much easily.

First note that $F_0\equiv0\equiv8\equiv F_6\pmod{4}$, and that $F_1\equiv1\equiv13\equiv F_7\pmod{4}$. One can then easily prove by induction that $$F_n\equiv F_{n+6}\pmod{4}.$$ Therefore, $$F_{1001}\equiv F_{6\cdot166+5}\equiv F_5\equiv5\equiv1\pmod{4},$$as we wanted.

ViHdzP
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  • Sorry to not answer your question directly, but it's the best that is possible. Even heavy-duty machinery like Wolfram|Alpha is incapable of factoring such a big number. In general, factoring is mostly useless when trying to solve congruences. In this particular case, even just working out the number itself is easier. – ViHdzP Oct 28 '19 at 02:09
  • Thank you so much I figured out that since if $F_n$ is equal to $1\mod4$ then all the prime divisor is equal to $1\mod4$. So as long as we found a prime divisor that is equal to $1\mod4$ then we passively prove that $F_n$ is equal to $1\mod4$ – Algorisum Oct 28 '19 at 02:11
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    Actually, if you already know that all prime factors of $F_{1001}$ are congruent to 1 mod 4, you can easily figure out that their product (namely $F_{1001}$) must also be congruent to 1 mod 4. In general, if $a\equiv c\pmod{n}$ and $b\equiv d\pmod{n}$, then $ab\equiv cd\pmod{n}$. Don't know how you figured this out, though, this seems like slight overkill. – ViHdzP Oct 28 '19 at 02:18
  • Thats because if we figured out a prime divisor then all the prime number multiply by the prime divisor is equal to $1\mod4$ – Algorisum Oct 28 '19 at 02:26
  • Hello again, may I ask how you got $F_{6\cdot166 + 5} \equiv F_5$? – Algorisum Oct 28 '19 at 03:03
  • But a number can be congruent to 1 mod 4 even if not all of its prime factors are. Consider $21=3\times7$. Even so, after checking the first few cases with Wolfram|Alpha, it does look like $F_n$ can only have prime factors of the form $4k+1$ for odd $n$, but proving this seems far from trivial. – ViHdzP Oct 28 '19 at 05:12
  • By what we proved, we have that $F_5\equiv F_{6\cdot1+5}\equiv F_{6\cdot2+5}\equiv\ldots\equiv F_{6\cdot166+5}\pmod{4}$. In general, knowing already that $F_n\equiv F_{n+6}\pmod{4}$, another induction easily proves that $F_n\equiv F_{n+6k}$. – ViHdzP Oct 28 '19 at 05:16
  • By the way, here’s the proof that all odd prime factors of $F_n$ are of the form $4k+1$ for odd $n$, adapted from https://www.fq.math.ca/Scanned/8-1/daykin-a.pdf. Suppose that $p\mid F_n$. Since $F_{n-1}^2-F_nF_{n-2}=(-1)^n=-1$, $F_{n-1}^2\equiv-1\pmod{p}$. However, it is well known that this is possible only for primes of precisely the form we want. For more info on that last fact, you might want to read up on quadratic residues and Euler’s Criterion. – ViHdzP Oct 28 '19 at 06:09
  • Thank you for your clarification – Algorisum Oct 28 '19 at 23:01
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It is actually possible to factor $F_{1001}$.

\begin{align*} F_{1001}=&13^2\cdot89\cdot233\cdot8009\cdot8581\cdot741469\cdot988681\cdot4832521\cdot159607993\\ &\cdot1929584153756850496621\cdot5811794973846976755532222929865278366042132879433\\ &\cdot46051361876019993056032153159112764928518076136032760569320871268374299454861127226737142167112433 \end{align*}

We have the factorizations \begin{align*} F_7&=13,\\ F_{11}&=89,\\ F_{13}&=233,\\ F_{77}&=13\cdot89\cdot988681\cdot4832521,\\ F_{91}&=13^2\cdot233\cdot741469\cdot159607993,\\ F_{143}&=89\cdot233\cdot8581\cdot1929584153756850496621. \end{align*} To factor $F_{1001}$, we first divide out by the primes that we have already found: $$F_{1001}=13^2\cdot89\cdot233\cdot8581\cdot741469\cdot988681\cdot4832521\cdot159607993\cdot1929584153756850496621\cdot C$$ where $C$ denotes a $151$-digit composite number. Suppose that $p$ is a prime number dividing $C$. Then the Fibonacci entry point $a_p$ must equal $1001$. One useful fact about Fibonacci entry points is that if $p\equiv1,4\pmod{5}$ then $a_p\bigm|p-1$ and if $p\equiv2,3\pmod{5}$ then $a_p\bigm|p+1$.

By the Chinese remainder theorem, $p$ has a remainder of $1$ or $3002$ or $3004$ or $4003$ modulo $5005$. We can do a little better by noting that $p$ must be congruent to 1 modulo 4 (by URL's comment on his answer). Then $p$ has a remainder of $1$ or $8009$ or $14013$ or $18017$ modulo $20020$. This immediately finds the prime factor $8009$ of $C$. Factoring the rest of $C$ is more challenging and is best done with the ECM.

Thomas Browning
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