Математикадан «Туймаада» олимпиадасы. Жоғары лига. 2013 жыл

${{p}^{2}}-pq-{{q}^{3}}=1$ теңдеуін жай сандар үшін шешіңіз. ( А. Голованов )

Комментарий/решение

Lemma:

if $p=3k+2$, then $p$ can not divide $q^2-q+1$.

Solution: $p|q^3+1$ then by fermat we get a contradiction.

Step 1)

Easily see that $p>3$, then $p^2=pq+q^3+1>2q+q^3+1>2q+q^2+1=(p+1)^2$,

hence $p>q+1$, then $p|q^3+1=(q+1)(q^2-q+1)$, and $q+1<p$, from this see that

$p|q^2-q+1$, hence $p=3k+1$.

Step 2)

if $q=3m+1$, then $1=p^2-pq-q^3 \equiv -1 (mod$ $3)$, a contradiction. Hence $q=3m+2$ or $q=3$. if $q=3m+2$, then $1=p^2-pq -q^3 \equiv0 (mod 3)$, conradiction again. Hence $q=3$, and from this we easily get $p=7$. So$(p,q)=(7,3)$ is the only solution.

Рассмотрим mod p, тогда: $p \mid q^3+1 , (q+1)(q^2-q+1) : p$, если $q+1 : p, \Rightarrow p^2 = pq + q^3 + 1 > q^2 \Rightarrow p^2 > q^2, p > q, p \ge q+1$, but $q+1 \ge p$ so contradicts,

esli $p = q + 1$, to $(q+1)^2 = q(q+1) = q^3 + 1, \Rightarrow q+1 = q^2 + 1 > q +1$ opyat contradicts, so $q^2-q+1 : p$, esli $p = 2(mod 3)$, then $q^2-q+1=3k+2 = 2(mod 3)$ contraditcs, if $p = 0 (mod 3), p = 3, q(q^2+3) = 8$ no solution, so $p = 1(mod3), q^3+q = 0 (mod 3)$, so $q=3, p^2 = 3p + 28, 28 : p$ tut $p = {2, 7}$ otkuda edinstvenniy solution $p = 7, q = 3$