0

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.

misi10
  • 107
  • Are you asking if those equations factor mod n? Do you know about finite fields? – Zach L. Dec 04 '12 at 12:48
  • I want to get roots and i don't know anything about finite fields? – misi10 Dec 04 '12 at 13:02
  • Do keep in mind that neither $25$ nor $49$ is a prime, so the product of two non-zero things can be zero in those rings. Another thing is that there is always "completing the square" trick, so the latter polynomial $$x^2+4x+2=x^2+4x+4−2=(x+2)^2−2\equiv(x+2)^2−100 \pmod{49}$$ can be written as a difference of two squares and hence factored. With the first polynomial I would just brute force find its zeros modulo $5$, and then check whether these come from roots modulo $25$. Remember that if $a$ is a root modulo $25$, then it's also a root modulo $5$. – Jyrki Lahtonen Dec 04 '12 at 13:25
  • If a is a root of modulo 5 then is's also a root module 5? can you explain more 1) ? – misi10 Dec 04 '12 at 13:39
  • Following @JyrkiLahtonen's hint: $,a^2-b^2=(a-b)(a+b),$ , which is true modulo any natural... – DonAntonio Dec 04 '12 at 13:42
  • @misi10: For example if you were to solve the equation $x^3-1=0\pmod{25}$ you first look at it modulo $5$. You will quickly find that for $x^3-1$ to be divisible by $5$ we must have $x\equiv1\pmod5$. Now if $25\mid x^3-1$, then surely $5\mid x^3-1$, so $x$ must be congruent to one of $1,6,11,16,21$ modulo $25$. More trial-and-error... (at least until you get into $p$-adics, which won't usually happen for a year or three). – Jyrki Lahtonen Dec 04 '12 at 15:08
  • Please double check that if the first equation is $\rm:2x^\color{#C00}3+,\ldots$ vs. $\rm:2x^\color{#C00}2+,\ldots\ .\ \ $ – Bill Dubuque Dec 04 '12 at 15:28

1 Answers1

0

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 $\rm\ 5^2\!\mid (x+4)(2x-1).$

Community
  • 1
Bill Dubuque
  • 272,048