Show that for all prime numbers $p$ greater than $3$, $24$ divides $p^2-1$ evenly.

Question

Show that for all prime numbers $p$ greater than $3$, $24$ divides $p^2-1$ evenly.

Since $(p+1)(p-1) = p^2-1$ we have that $\frac{(p+1)(p-1)}{24}=k$, where $k \in \Bbb Z.$

Now since $24 = 2^3 \cdot 3$ and the numerator contains always at least one even factor(?) we have that $24=2^3\cdot3\vert(p+1)(p-1).$

Is my reasoning here correct or am I missing something here?

You could conclude that from the fact that $\binom{p+1}{3}$, $\binom{p+1}{4}$ are integers and $1\cdot 2\cdot 3\cdot 4 = 24$. — rtybase, Oct 24 '20 at 14:19
Does this answer your question? For any prime $p > 3$, why is $p^2-1$ always divisible by 24?. Also, there's If $p$ is prime, with $p> 3$, then $p^{2} \equiv1\pmod{24}$ and several others. — John Omielan, Oct 24 '20 at 18:18

Albus Dumbledore · Answer 1 · 2020-10-24T14:30:11.693

1

No its wrong you have not shown $2^3,3$ divides $p^2-1$.

For the correct proof use the hint given below.

Hint: use the fact that any prime $p\ge 5$ is of form $6k+1$ or $6k-1$

edited Oct 24 '20 at 14:30

answered Oct 24 '20 at 14:02

Albus Dumbledore

Well $p=5$ is also in the form $6k+5$ – player3236 Oct 24 '20 at 14:04
@player3236 ah yes i heve edited please see – Albus Dumbledore Oct 24 '20 at 14:07
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My point is you don't need to treat the case $k=0$ separately. Alternatively, using the form $6k \pm 1$ will simplify the proof, and you get to keep your "natural numbers". – player3236 Oct 24 '20 at 14:09

Lion Heart · Answer 2 · 2020-10-24T14:15:00.523

Let $p$ be a prime number one of the following forms and $k\in Z^+$.

If $p=3k+1$

$$p^2-1= (p-1)(p+1)=(3k+1-1)(3k+1+1)=3k(3k+2)$$

Or If $p=3k+2$

$$p^2-1= (p-1)(p+1)=(3k+2-1)(3k+2+1)=(3k+1)(3k+3)=3(3k+1)(k+1)$$

Hence $p$ can divisible by $3$

If $p=4k+1$

$$p^2-1= (p-1)(p+1)=(4k+1-1)(4k+1+1)=4k(4k+2)=8k(2k+1)$$

Or If $p=4k+3$

$$p^2-1= (p-1)(p+1)=(4k+3-1)(4k+3+1)=(4k+2)(4k+4)=8(2k+1)(k+1)$$

Hence $p$ can divisible by $8$

Therefore $p$ can divisible by $24$

score 1 · Answer 3 · answered Oct 24 '20 at 14:08

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When doing these kinds of problems it's best to avoid division - in this case, attempting to divide by 24.

Hints:

Since $p$ is odd, you have good information about the parity of $p\pm 1$ . One is even and the other doubly even.

Among three consecutive integers exactly one is divisible by $3$, and $p$ itself is not.

answered Oct 24 '20 at 14:08

Ethan Bolker

Hmm, so if I consider $\pmod{2}$ would I have that $p^2 - 1 \equiv 0 \pmod{2}$ since two even numbers multiplied gives an even number? – Oct 24 '20 at 14:12
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Sure, Daniel, but that's not enough. You have to show that $p^2 - 1 \equiv 0 \pmod{8}$. And also $p^2 - 1 \equiv 0 \pmod{3}$ – PM 2Ring Oct 24 '20 at 14:16
Would the fact that all even numbers are either $0,1$ or $2$ $\pmod{3}$ help here? – Oct 24 '20 at 14:19
@Daniel Keep the reasoning for $2$ and $3$ separate. For $3$ you want to think about $p-1, 0, p+1$. For $2$ you think about $p-1$ and $p+1$. You know those are both even. Explain why one of them must be doubly even. – Ethan Bolker Oct 24 '20 at 15:55

3 Answers3