(n^3 -n)(5^{8n+4} +3^{4n+2}) is a multiple of 3804

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(n^3 -n)(5^{8n+4} +3^{4n+2}) is a multiple of 3804

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Source: 1st Mexican Mathematical Olympiad 1987 OMM P6

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#1 h Jul 27, 2018, 8:55 PM • 2 Y

Y by Adventure10, Mango247

Prove that for every positive integer n the number $(n^3 -n)(5^{8n+4} +3^{4n+2})$ is a multiple of $3804$ .

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#2 h Jul 27, 2018, 10:16 PM • 2 Y

Y by Adventure10, Mango247

Let $M=(n^3 - n)(5^{8n+4} +3^{4n+2})$ . Clearly $12\mid M$ , so it suffices to prove $317\mid M$ .
For this, note that $5^4\equiv -9\pmod{317}$ , so $5^{8n+4} +3^{4n+2}\equiv (-9)^{2n+1} + 9^{2n+1}\equiv 0\pmod{317}$ , since $2n+1$ is always odd. We are done.

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#3 h May 17, 2022, 5:46 AM

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Solution

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#4 h Mar 26, 2025, 9:03 AM

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The expression is equal to $n(n-1)(n+1)(625^{2n+1}+9^{2n+1})$ . As in three consecutive numbers there is always a multiple and at least one pair $6\mid n(n-1)(n+1)$ . It is clear that $2n+1$ is odd, so $625+9\mid 625^{2n+1}+9^{2n+1}$ and $625+9=634$ . So $6\cdot 634=3804$ divides $(n^3-n)(625^{2n+1}+9^{2n+1})$ .

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