Mathematical and Physical Journal
for High Schools
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Problem B. 4365. (May 2011)

B. 4365. Find all positive integers n such that 2n-1 and 2n+2-1 are both primes, and 2n+1-1 is not divisible by 7.

(Suggested by S. Kiss, Budapest)

(3 pont)

Deadline expired on June 10, 2011.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Ha \(\displaystyle k=1\), akkor \(\displaystyle 2^k-1=1\) nem prím. Ha \(\displaystyle k=ab\), ahol \(\displaystyle a,b\) 1-nél nagyobb egész számok, akkor \(\displaystyle 2^a-1>1\) valódi osztója a \(\displaystyle 2^k-1\) számnak. Ezért ha \(\displaystyle 2^n-1\) és \(\displaystyle 2^{n+2}-1\) is prím, akkor \(\displaystyle n\) és \(\displaystyle n+2\) is prím kell legyen. Mármost ha \(\displaystyle n>3\), akkor \(\displaystyle n\) páratlan, tehát \(\displaystyle n+1\) páros. Továbbásem \(\displaystyle n\), sem \(\displaystyle n+2\) nem osztható 3-mal, tehát \(\displaystyle n+1\) osztható 3-mal, így 6-tal is. Ekkor viszont \(\displaystyle 2^6-1\mid 2^{n+1}-1\) miatt \(\displaystyle 2^{n+1}-1\) osztható \(\displaystyle 7\)-tel. Mivel \(\displaystyle n=2\) esetén \(\displaystyle n+2\) nem prím, az egyetlen lehetőség \(\displaystyle n=3\), ami meg is felel a feltételeknek.


78 students sent a solution.
3 points:56 students.
2 points:15 students.
1 point:5 students.
0 point:1 student.
Unfair, not evaluated:1 solution.

Problems in Mathematics of KöMaL, May 2011