How to find all integer $x$ such that $x^{2}+5145x+2332\equiv 0\pmod{ 9797}$. I think $9797$ can factor, then $9797=97\cdot11$. So I can convert to $$x^{2}+5145x+2332\equiv 0\pmod{97}$$ $$x^{2}+5145x+2332\equiv 0\pmod{11}$$ How do to next? Thanks.
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2$9797=97\times 101$. – player3236 Nov 27 '20 at 05:38
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3It's $x^2+4x+4\equiv0\bmod97$ and $x^2-6x+9\equiv0\bmod101$. Do you know the Chinese remainder theorem? – J. W. Tanner Nov 27 '20 at 05:45
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@JWTanner why do I see the misspelling "proove" so often? – Parcly Taxel Nov 27 '20 at 06:01
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@ParclyTaxel: perhaps people get confused because proof has two os – J. W. Tanner Nov 27 '20 at 06:03
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You can ese the same CRT-based method as in the linked dupe. – Bill Dubuque Nov 27 '20 at 08:40
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The coefficients may be factored as follows: $$x^2+(15×7^3)x+(44×53)\equiv0\bmod(97×101)$$ Modulo $97$, the equation becomes $$x^2+4x+4\equiv(x+2)^2\equiv0\bmod97$$ Modulo $101$, $$x^2-6x+9\equiv(x-3)^2\equiv0\bmod101$$ (We have used the smallest absolute value possible for calculations to simplify the work.) Thus $x\equiv-2\bmod97\equiv3\bmod101$, or by the Chinese remainder theorem $x\equiv2326\bmod9797$.
Parcly Taxel
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