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Find all natural number $n$ such $n^{6}-5n^{4}+3n^{2}+1$ is divisible by $399$.

My try; Let be $f\left( n \right)=n^{6}-5n^{4}+3n^{2}+1, n\in \mathbb{N}$. Since $399=3\cdot7\cdot19\cdot$ and $3,7,19$ are prime number then therefore 399|$f\left( n \right)\Longleftrightarrow 3,7,19|f\left( n \right)$. Then from the algorithm of dividing the number n by the numbers $3,7$ and $19$ ,we have
$n=3q_{1}+r_{1}$ where $0\le r_{1}\lt 3$, $n=7q_{2}+r_{2},$ $0\le r_{2}\lt 7$ and $n=19q_{3}+r_{3}$ , $0\le r_{3}\lt 19$ . Maybe now by analyzing these cases ,it will help us to find the required numbers ?

If you have any tips for this or you have any other solution , I would be very grateful if you share it with us .

Bill Dubuque
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Fsinani
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    Test by brute force which values of $r_1,r_2,r_3$ are solutions, and combine them using the Chinese remainder theorem. – Anne Bauval Dec 26 '23 at 22:32
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    It may be useful to factor $f(n)$. – Travis Willse Dec 26 '23 at 22:35
  • Furthermore $ f(n) \equiv f(-n)$, so you only need to test half of those cases. So ~15 values to check, which isn't too bad. – Calvin Lin Dec 26 '23 at 23:46
  • $\pm1$ is a root so $f(n)=(\color{#0a0}{n^2-1})(n^4-4n^2-1) = (n^2-1)g(n^2).,$ $g(n)$ has no roots mod $3$ and $7$ since its discriminant $20$ in't square, but $!!\bmod 19!:\ 20\equiv 1^2$ so $g(n^2) = (n^2!+!7)(n^2!+!8)$ $ = (n^2!+!7)(n^2!-!\color{#c00}7^2)$. Finally CRT combine & lift the roots $\color{#0a0}{\pm1}\bmod 3,7,19,$ and $\color{#c00}{\pm7}\bmod 19$ as here in the linked dupe. $\ \ $ – Bill Dubuque Dec 26 '23 at 23:56
  • Thank you very much for your tips , I appreciate all your tips. I'm still working to find up the solution . But i don't think my problem should be as a "duplicate" or an closed problem . – Fsinani Dec 27 '23 at 00:19
  • The CRT-based method is the same for any polynomial. The only work left is to combine & lift the roots by CRT - which is a rote mechanical procedure (cf. linked dupes). If you get stuck then post further comments and we can elaborate. – Bill Dubuque Dec 27 '23 at 00:38

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