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If possible, solve $3x^2 + 6x + 5 \equiv 0 $(mod 539)

  • I am struggling with several problems just like the one above, and decided to post what I believe is the hardest one to hopefully able to do the the rest of them. My professor's hint on it was to keep in mind that $539 = 11 * 49$ This wasn't much help unfortunately, any help is greatly appreciated.
Nick Powers
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  • Can you solve the equation modulo 11 and modulo 49? My guess is that the Chinese Remainder Theorem is where that hint was aimed at. – Ken Duna Mar 21 '16 at 19:02
  • Note that you can use the quadratic formula e.g. modulo $11$ so that $b^2-4ac=36-60=-24\equiv 9 \bmod 11$ and $-24\equiv 25 \bmod 49$ giving obvious squares whose square roots can be plugged in. – Mark Bennet Mar 21 '16 at 19:08
  • Usually people who keep asking questions within the same subtopic gradually learn something from the answers to their previous questions. – Bob Happ Mar 21 '16 at 21:14

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It is enough to solve mod $11$ and mod $49$, and then combine the results using the Chinese Remainder Theorem. There are two solutions mod $11$, which you can find without too much trouble by trying the $11$ possibilities $x = 0,1,\ldots, 10$. There are also two solutions mod $49$. Combining them, you should get $4$ solutions in all.

$49 = 7^2$, so you can first solve mod $7$ (again obtaining two possibilities), and then further examine each of those. Thus one of the solutions mod $7$ is $x\equiv 1 \mod 7$. If $x = 1 + 7 y$, $$3x^2 + 6 x + 5 = 147 y^2 + 84 y + 14 \equiv 7 (5 y + 2) \mod 49$$ so you need $5y + 2 \equiv 0 \mod 7$, and thus $y \equiv 1 \mod 7$ so $x \equiv 8 \mod 49$.

Robert Israel
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