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Question :- $A=k^2-1$ and $B=(k+1)^2-1$, where k is a natural number greater than 1. How many prime numbers are there by which both A and B are divisible for at least 1 value of k ?

I attempted this question by plugging different kinds of numbers from the families of primes into k

For instance, k to be a one of the number from Family of $2$ ($2n,2n+1$) then from Family of $3$ ($3n,3n+1,3n+2$)

For $k=3n+1$, I was able to get both A and B a multiple of $3$ , hence by hit and trial I confirmed $3$ to be one of the prime number by which A and B will be divisible, with k being of type $3n+1$

Now for how many prime numbers do I need to do the same thing $5$ $(5n+1,5n+2,5n+3,5n+4)$,$7$ $( 7n+1,7n+2,7n+3,7n+4,7n+5,7n+6)...$ ? How to solve the question efficiently?

Fin27
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  • Have you tried any examples? I think I quick search should tell you what $\gcd(A,B)$ must be. Of course, you'll then have to prove that. – lulu Mar 23 '23 at 16:48
  • gcd $((k-1)(k+1), k(k+2))$ = gcd$(k-1,k+2)$ = gcd$(k-1,3)$ , is it correct ? I don't know what would be the gcd of this final expression , and how would gcd help us in solving this question ? – Fin27 Mar 23 '23 at 16:54
  • Well, if the gcd is a divisor of $3$ then you have just eliminated every prime except $3$. – lulu Mar 23 '23 at 16:58
  • Thank you @lulu, but I am not sure if the expression of $gcd(k−1,k+2)$ is correct or not, could you please check if my reasoning for coming up with that expression is right or wrong :- Since, gcd$(k,k+1)=1$ , so $k+1$ will not contribute to the gcd , hence my updated expression will become gcd$(k-1,k*(k+3))$, then similarly gcd$(k−1,k)=1$, hence $k$ will not contribute to the gcd, therefore we would finally be left with gcd$(k−1,k+2)$ ? – Fin27 Mar 23 '23 at 17:19
  • Oh, yes. You are correct. The gcd is 3 if k=3n+1, and 1 otherwise – lulu Mar 23 '23 at 18:20
  • Euclid's Lemma & $\color{#c00}{(k,k^2!-!1)!=!1}\Rightarrow $ $(\color{#c00}{k^2!-!1,k}(k!+!2)) = (\color{#0a0}{k^2!-!1},\color{#90f}{k!+!2}) \overset{\rm\color{#0a0}R}= (\color{#0a0}3,k!+!2),,$ where the final equality follows by employing $\rm\color{#0a0}{R}$ = gcd mod reduction, using $\bmod \color{#90f}{k!+!2}!:\ k\equiv -2\Rightarrow \color{#0a0}{k^2!-!1\equiv 3}$. Finally the gcd $,(3,k!+!2)=3$ if $,k\equiv 1\pmod{!3},,$ else the gcd $=1\ \ $ – Bill Dubuque Mar 23 '23 at 23:57
  • Re: proof in your prior comment: to remove the factor $k+1$ from the first arg of the gcd we need it to be coprime to the 2nd arg, i.e. to both $k$ and $k+2,,$ but you didn't mention such for $k+2$. Once you do so the proof is fine. We can also use gcd mod reduction instead of your second invocation of Euclid's Lemma, i.e. $(k-1,f(k)) = (k-1,f(1))$ for a polynomial $,f(x)\in\Bbb Z[x]$ (as in my prior comment). – Bill Dubuque Mar 24 '23 at 00:10
  • Once all is clear its best to delete this question since we already have too many questions on this topic. – Bill Dubuque Mar 24 '23 at 00:16

1 Answers1

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Note that you have

  • $A=k^2-1=(k+1)(k-1)$, and
  • $B=(k+1)^2-1=k(k+2)$.

Thus:

  • $A$ and $B$ cannot be simultaneously even;
  • If $3$ divides $k-1$ then $3$ divides both $A$ and $B$;
  • every prime $p>3$ can divide at most one of $4$ consecutive numbers, so $p$ cannot divide $A$ and $B$ at the same time.
Andrea Mori
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  • Thanks Andrea, how did you come up with the following conclusion :- every prime >3 can divide at most one of 4 consecutive numbers ? – Fin27 Mar 23 '23 at 17:07
  • @Fin27: it's obvious. If $p$ is a prime the only nonnegative integers divisible by $p$ are $0$, $p$, $2p$, $3p$ and so on. They are separated by intervals of length $p$. – Andrea Mori Mar 23 '23 at 17:11