Problem E:17 120 - celebrating my 65th birthday

 

(On the 65th anniversary: may-08-2025)

         " E:17120.  Determine all prime numbers $n$ that have the property that the

           numbers $n^2+6$ and $n^2-6$ are prime .

(Emil Gheorghiu, Timișoara)"

ANSWER CiP

$$n=5$$

                    Solution CiP

                Let's make some attempts:

$n=2\;\;:\;n^2+6=10-this\; is \;NOT \;prime,\;\;n^2-6=-2;$

$n=3\;\;:\;n^2+6=15-this\; is \;NOT \;prime,\;\;n^2-6=3;$

$n=5\;\;:\;n^2+6=31,\;\;n^2-6=19 \;-BOTH \;are\; prime\;\;;$

$n=7\;\;:\;n^2+6=55-this\; is \;NOT \;prime,\;\;n^2-6=43;$

$n=11\;\;:\;n^2+5=127,\;\;n^2-6=115-this\; is \;NOT \;prime:$

$n=13\;\;:n^2+6=175-this\; is \;NOT \;prime,\;\;n^2-6=163;\dots$

     Moreover, it is observed that for $n\geqslant 7$, one of the sought numbers is a multiple of 5.

          Any natural number has one of the forms $5k,\; 5k\pm1,\;5k\pm2,\;\;k\in \mathbb{N}$. 

          If $n=5k$, being prime it can only be 5. The situation is convenient.

          If $n=5k\pm1$ then 

$n^2-6=(25k^2\pm10k+1)-6=25k^2\pm10k-5=5\cdot (5k^2\pm2k-1)$ 

so it is a multiple of 5 and greater than 5 for $n\geqslant11$. None of the numbers are convenient.

          If  $n=5k\pm2$ then 

$n^2+6=(25k^2\pm20k+4)+6=25k^2\pm20k+10=5 \cdot (5k^2\pm4k+2)$

so it is a multiple of 5 and greater than 5 for $n\geqslant 7$ . None of the numbers are convenient.

          So the only prime number $n$ with the desired property is 5.

$\blacksquare$

          Remark CiP It was an attempt, without much success, to consider the cases $n=6k\pm1,\;\;k\in \mathbb{N}$, the only forms that give prime numbers $\geqslant 5$ and cases $n=2$ and $n=3$ should be treated separately.

     We are left with only the conclusion: 

Always one of the four numbers $12k(3k\pm1),\;\;12k(3k\pm1)+2$ is divisible by 5.

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