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

luni, 7 septembrie 2020

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 $m$;

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 $d_{295}=1$

ogeometrie