Problem MA 82 CRUX MATHEMATICORUM vol 46, no 7 pag 285

 

                    ANSWER CP : $\underset{\;\;n}max\;d_{n}=197$

 More precisely    $d_{197\cdot k -100}=197$ , $d_{n}=1\;\;if\;n\neq 197\cdot k-100$

Solution CP

      We apply the following properties:

(A)       $(a,b)=(a,-b)$;

(B)        $(a,b)=(a+b\cdot m,b)=(a,b+a \cdot m)$ , for any integers $m$;

(C)        if $(c,b)=1$ then $(a,b)=(a \cdot c,b)$.

Here $(a,b)$ denote "greatest common divisor" for integers $a$ and $b$. See for (C) an ancient post.

      Let $d_{n}=(a_{n},a_{n+1})=(n^{2}+2\cdot n+50, (n+1)^{2}+2(n+1)+50)=$

        $=(n^{2}+2\cdot n+50,n^{2}+4\cdot n+53)=$

        $\overset{(B)}=(n^{2}+2\cdot n+50,n^{2}-4\cdot n+53-(n^{2}+2\cdot n+50))=$

     $=(n^{2}+2\cdot n+50,2\cdot n+3)\;\overset{(C)}{=}\underset{(2,2n+3)=1}{=}{=}(2(n^{2}+2n+50),2n+3)=$

     $=((2n+3)n+n+100,2n+3)\overset{(B)}{=}(n+100,2n+3)\overset{(B)}{=}(n+100,2n+3-2(n+100))=$

     $\overset{(A)}{=}(n+100,197)$.

The number 197 being prime

             if $n+100=197\cdot k$ then $d_{n}=197$

             and if $n+100=197\cdot k+r$, $r=1, 2, ... ,196$ then $d_{n}=1$.

$\blacksquare$

          Examples

     $a_{96}=5458=2\cdot 4729;$

      $a_{97}=9653=7^{2}\cdot 197$,    hence $d_{96}=1$

      $a_{98}=9850=2\cdot 5^{2}\cdot 197$,    hence $d_{97}=197$

      $a_{99}=10\; 049=13 \cdot 773$,     hence $d_{98}=1$

....................................................................................................

      $a_{293}=86\;485=5\cdot 7^{2}\cdot 353$

      $a_{294}=87\;074=2\cdot 13\cdot17\cdot 197$,     hence $d_{293}=1$

      $a_{295}=87\;665=5\cdot 89\cdot 197$,      hence $d_{294}=197$

      $a_{296}=88\;258=2\cdot 44\;129$,       hence $d_{295}=1$