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For integer $n\ge2$,
Is there always a prime p such that $v_p(n^3-n)=1$?
For example, n=2$n=2$: p=2(6=2×3)$p=2$$(6=2\times3)$ n=3$n=3$: p=3(24=2³×3)$p=3$$(24=2^3\times3)$ n=9$n=9$: p=5(720=2⁴×3²×5)$p=5$$(720=2^4\times 3^2\times5)$
For integer $n\ge2$,
Is there always a prime p such that $v_p(n^3-n)=1$?
For example, n=2: p=2(6=2×3) n=3: p=3(24=2³×3) n=9: p=5(720=2⁴×3²×5)
For integer $n\ge2$,
Is there always a prime p such that $v_p(n^3-n)=1$?
For example, $n=2$: $p=2$$(6=2\times3)$ $n=3$: $p=3$$(24=2^3\times3)$ $n=9$: $p=5$$(720=2^4\times 3^2\times5)$
