Again, please help. Find all possible x such that x^4+2x^3+8x+9=0(mod 35).?

Since 5 * 7 = 35 and gcd(5, 7) = 1, it suffices to solve

x^4 + 2x^3 + 8x + 9 = 0 (mod 5) and x^4 + 2x^3 + 8x + 9 = 0 (mod 7).

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(i) x^4 + 2x^3 + 8x + 9 = 0 (mod 5)

Note that x = 0 (mod 5) is not a solution; hence we can say that x^4 = 1 (mod 5) by Fermat's Little Theorem.

==> 1 + 2x^3 + 8x + 9 = 0 (mod 5)

==> 2x^3 + 8x = 0 (mod 5), since 10 = 0 (mod 5)

==> 2x(x^2 + 4) = 0 (mod 5)

==> x^2 + 4 = 0 (mod 5), since we ruled out x = 0 (mod 5) as a solution

==> x^2 - 1 = 0 (mod 5), since 4 = -1 (mod 5)

==> (x + 1)(x - 1) = 0 (mod 5)

==> x = -1, 1 (mod 5), since 5 is prime.

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(ii) x^4 + 2x^3 + 8x + 9 = 0 (mod 7)

==> x^4 + 2x^3 + x + 2 = 0 (mod 7), reducing the coefficients mod 7

==> x^3 (x + 2) + 1(x + 2) = 0 (mod 7)

==> (x^3 + 1)(x + 2) = 0 (mod 7)

==> (x^3 + 8)(x + 2) = 0 (mod 7)

==> (x + 2)(x^2 - 2x + 4) * (x + 2) = 0 (mod 7), via sum of cubes

==> (x + 2)^2 (x^2 - 2x - 3) = 0 (mod 7)

==> (x + 2)^2 * (x - 3)(x + 1) = 0 (mod 7)

==> x = -2, 3, or -1 (mod 7), since 7 is prime.

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In summary, we have that

x = -1, 1 (mod 5) and x = -2, 3, or -1 (mod 7).

<==> x = 1 or 4 (mod 5) and x = 3, 5, or 6 (mod 7).

By the Chinese Remainder Theorem (or otherwise), we find that we have two solutions modulo 35:

(1) x = 1 (mod 5) and x = 3 (mod 7) ==> x = 31 (mod 35)

(2) x = 1 (mod 5) and x = 5 (mod 7) ==> x = 26 (mod 35)

(3) x = 1 (mod 5) and x = 6 (mod 7) ==> x = 6 (mod 35)

(4) x = 4 (mod 5) and x = 3 (mod 7) ==> x = 24 (mod 35)

(5) x = 4 (mod 5) and x = 5 (mod 7) ==> x = 19 (mod 35)

(6) x = 4 (mod 5) and x = 6 (mod 7) ==> x = 34 (mod 35)

That is, x = 6, 19, 24, 26, 31, or 34 (mod 35).

(Double checked on Wolfram Alpha.)

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I hope this helps!

I had tried the problem later and luckily success:

35 could be splited into 7 *5=>there were 2 equations:

x^4+2x^3+8x+9=0(mod 7)-------(1)

and

x^4+2x^3+8x+9=0(mod 5)-------(2)

From (1), by trial and error, got

x=3,5,6 (mod 7)--------(3)

From (2), got

x=1,4 (mod 5)---------(4)

In order to find all x satisfying both (3) & (4),

I considered these combinations:

x=3 (mod 7) & x=1 (mod 5)------(5)

x=3 (mod 7) & x=4 (mod 5)------(6)

x=5 (mod 7) & x=1 (mod 5)------(7)

x=5 (mod 7) & x=4 (mod 5)------(8)

x=6 (mod 7) & x=1 (mod 5)------(9)

x=6 (mod 7) & x=4 (mod 5)------(10)

For (5)

5x=15 (mod 35)

7x=7 (mod 35)

=>

2x=-8(mod 35)

=>

x=-4(mod 35)

=>

x=31 (mod 35)

Following the similar methods, I finally got

x=6, 19, 24, 26, 31, 34 (mod 35) or

x=6+35k

x=19+35k

x=24+35k

x=26+35k

x=31+35k

x=34+35k

where k is an integer.

Thank you all to provide other

good ways for the solution.

Let T = x^4 + 2x^3 + 8x + 9 and using calculator for trial substitutions

x = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} found first few integer values as

T/7 = {f, f, 24, f, 132, 255, f, f, f, 1727, f} (f rejected fraction values)

T/5 = {4, f, f, 85, f, 357, f, f, 1620, f, 3480}

255 *7 = 357*5 = 1785 when x = 6

So, x = 6 gave x^4 + 2x^3 + 8x + 9 = 51*35 = 0 (mod 35)

T/5 values correspond to x = 5a + 1 or (5a + 1) + 3 so for x < 35 these are:-

1, 6, 11, 16, 21, 26, 31, and

4, 9, 14, 19, 24, 29, 34

Extending the list for T/7 you can eventually convince yourself that

T/7 values correspond to x = 5b + 3 or (5b + 3) + 2 or (5b + 3) + 3, namely,

3, 10, 17, 24, 31, and

5, 12, 19, 26, 33, and

6, 13, 20, 27, 34,

The only values in both lists are x = 6, 19, 24, 26, 31, or 34

This is not a recommended method as the three forms for T/7 not obvious.

Use the brilliant method breaking down the polynomial asposted here by KB

Check values of x from 0 to 34 using a spreadsheet, substituting into the polynomial to look for zeros mod 35.

Results are

x=k=6, 19, 24, 26, 31, 34.

Solutions are x=35n+k where k is as above and n is any integer (n in Z).

Checked using Wolfram Alpha.