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How to decompose equations below and then solve

1)$ 2x^3 + 7x - 4 \equiv 0 \pmod{25} $

2)$ x^2+4x+2 \equiv 0 \pmod{49}$

Thank you.

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    Please double check that if the first equation is $\rm\:2x^\color{#C00}3+\,\ldots$ vs. $\rm\:2x^\color{#C00}2+\,\ldots\ .\ \ $2012-12-04

1 Answers 1

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Completing the square $\rm\ mod\ 49\!:\ x^2\!+4x+2 = (x+2)^2\!-2 \equiv (x+2)^2\! - 10^2 \equiv (x-8)(x+12).\:$ Therefore $\rm\: 7^2\mid (x-8)(x+12)\:\Rightarrow\: 7^2\mid x-8\ $ or $\rm\ 7^2\mid x+12\ $ or $\rm\ 7\mid x-8,\, x+12.\:$ The last case is impossible since it implies that $\rm\:mod\ 7\!:\,\ 8\equiv x\equiv -12.$

Quite likely, the first problem is a misprint for $\rm\:f = 2x^\color{#C00}2+7x-4.\:$ Using the AC method, with $\rm\: X = 2x,\:$ we have $\rm\:2f = X^2+7X-8 = (X+8)(X-1) = (2x+8)(2x-1),\:$ therefore $\rm\: f = (x+4)(2x-1).\:$ Now, as above, apply a case analysis to $\rm\ 5^2\!\mid (x+4)(2x-1).$

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    How to get more example with (modular arithmetic) and (equation)?2012-12-04