5 votos

¿Existen números primos distintos$p,q,r$ tales que$\frac{p+q+r}{3}; \frac{p+q}2; \frac{r+q}2; \frac{p+r}2$ son números primos.

¿Existen números primos distintos$p,q,r$ tales que$$\frac{p+q+r}{3}; \frac{p+q}2; \frac{r+q}2; \frac{p+r}2$ $ son números primos.

Mi trabajo:

Encontré $(p,q,r)=(5,17,29)$.

¿Hay algún otro ejemplo?

4voto

Anthony Cramp Puntos 126

Hay secuencias aritméticas arbitrariamente largas de primos. REFERENCIA
Deje $$ a_0

El mismo principio "Ramsey" nos permite encontrar muchos otros patrones finitos en el conjunto de primos.

3voto

The Short One Puntos 61

Si tiene Mathematica, puede obtener fácilmente cientos de ejemplos si no miles:

{{5,17,29},{5,17,89},{5,17,101},{5,17,197},{5,17,449},{5,17,521},{5,29,53},{5,29,89},{5,29, 269},{5,29,509},{5,41,137},{5,41,173},{5,41,257},{5,41,293},{5,53,461},{5,89,173},{5,89, 449},{5,101,113},{5,101,197},{5,113,353},{5,137,197},{5,137,449},{5,173,293},{5,173, 521},{5,197,269},{5,197,389},{5,197,521},{5,257,461},{5,257,509},{5,269,449},{5,293, 509},{5,353,449},{5,389,449},{7,19,67},{7,31,271},{7,67,127},{7,67,139},{7,67,379},{7, 79,127},{7,79,307},{7,127,439},{7,139,307},{7,211,271},{7,307,439},{11,23,503},{11,47, 71},{11,47,251},{11,71,131},{11,71,491},{11,83,443},{11,83,503},{11,107,191},{11,131, 167},{11,131,251},{11,131,347},{11,131,491},{11,263,443},{11,347,491},{13,61,193},{13,73, 181},{13,73,241},{13,109,349},{13,193,373},{13,241,433},{13,349,409},{17,41,101},{17,41, 461},{17,41,521},{17,101,461},{17,197,257},{17,197,509},{17,257,449},{19,43,439},{19,67, 127},{19,103,199},{19,103,379},{19,139,523},{19,199,283},{19,199,463},{19,283,379},{23, 59,239},{23,71,443},{23,83,311},{23,83,431},{23,191,503},{23,251,443},{29,53,509},{29,89, 173},{29,113,149},{29,113,269},{29,113,449},{29,149,233},{29,197,317},{29,269,509},{29, 353,449},{31,43,103},{31,43,163},{31,43,523},{31,103,283},{31,127,331},{31,163,223},{37, 97,157},{37,241,421},{41,53,173},{41,317,521},{43,79,379},{43,103,523},{43,151,523},{43, 163,283},{43,271,283},{43,379,499},{47,59,467},{47,71,131},{47,131,491},{47,167,179},{47, 167,419},{47,167,467},{47,179,347},{47,227,479},{47,311,491},{47,347,419},{47,419, 467},{53,89,449},{53,113,281},{53,173,293},{53,173,461},{53,449,509},{59,419,443},{59, 479,503},{61,241,421},{61,397,421},{67,79,487},{67,127,487},{67,379,487},{71,131, 431},{71,191,311},{71,227,491},{71,311,467},{71,431,491},{73,241,373},{73,349,409},{79, 139,283},{79,199,223},{79,307,367},{79,379,463},{83,131,383},{83,131,503},{83,191, 263},{83,191,443},{83,251,263},{83,251,383},{83,263,443},{83,263,503},{89,113,269},{89, 113,389},{89,269,449},{101,113,233},{101,197,281},{101,233,353},{101,257,521},{103,151, 463},{103,223,523},{103,283,463},{103,379,439},{107,167,227},{107,167,359},{107,227, 239},{107,227,479},{107,347,359},{109,277,337},{109,277,457},{109,313,349},{113,233, 353},{113,269,449},{113,389,449},{113,389,509},{131,227,491},{137,197,257},{137,197, 389},{137,257,449},{137,317,389},{139,307,487},{149,233,389},{151,211,271},{163,379, 499},{167,191,311},{167,227,359},{167,347,419},{173,281,353},{173,389,449},{193,433, 541},{197,257,269},{227,239,467},{227,251,311},{239,383,479},{241,421,457},{257,281, 521},{269,449,509},{277,337,397},{283,331,463},{317,461,521},{337,409,457},{359,419, 479},{389,449,509},{401,461,521},{419,479,503},{431,467,491}}

Voy a verificar al azar uno de estos:$p = 47, q = 419, r = 467$. Entonces los promedios son:$311, 233, 257, 443$.

Aquí está el código que utilicé: Select[Subsets[Prime[Range[100]], {3}], PrimeQ[(#[[1]] + #[[2]] + #[[3]])/3] && PrimeQ[(#[[1]] + #[[2]])/2] && PrimeQ[(#[[1]] + #[[3]])/2] && PrimeQ[(#[[2]] + #[[3]])/2] &] Si quieres más, simplemente cambia el 100 a 1000 o lo que sea.

1voto

dmay Puntos 415

Sí. Algunos ejemplos:

  • $(p,q,r)=(5, 17, 89)$;
  • $(p,q,r)=(7, 19, 67)$;
  • $(p,q,r)=(5, 29, 53)$;
  • $(p,q,r)=(11, 47, 71)$.

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