# need a easy way to solve it

• Sep 17th 2012, 09:23 AM
Petrus
need a easy way to solve it
For which integers are x^5+4x^4+3x+1 divisible by 5?
i have solved this prob and there is none but i did it on a "hard way"
i did make it to x(x^3(x+4)+3)+1 and try put like x to 1,2,3 etc.. is there any easy way to solve it?

• Sep 17th 2012, 05:28 PM
Wilmer
Re: need a easy way to solve it
Wolfram gives this as solution:
solve x&#94;5&#43;4&#42;x&#94;4&#43;3&#42;x&#43;1-5&#42;k&#61;0 for x - Wolfram|Alpha

What d'heck does it mean...Plato?
• Sep 17th 2012, 07:15 PM
Kiwi_Dave
Re: need a easy way to solve it
Let's plase all of the integers into one of 5 different classes:

x = 5a + 0, or
x = 5a + 1, or
x = 5a + 2, or
x = 5a + 3, or
x = 5a + 4

for each type of number I would try substituting for x in your original equation and determining if it is possible for the result to be divisible by 5. For example:

Let X = 5a + 0. Yor equation becomes:

$(5a)^5+4(5a)^4+3(5a)+1 = 5(5^4a^5+4.5^3.a^4+3a)+1$ and this cannot be divisible by 5 so the enture class of numbers x=5a+0 cannot give a result that is divisible by 5.

Let X = 5a + 1. Yor equation becomes:

$(5a+1)^5+4(5a+1)^4+3(5a+1)+1$

When you multiply out the first bracket only the last term is not divisible by 5, the last term is 1
When you multiply out the second bracket only the last term is not divisible by 5, the last term is 4
When you multiply out the third bracket only the last term is not divisible by 5, the last term is 3

so we can say that for some k

$(5a+1)^5+4(5a+1)^4+3(5a+1)+1 = 5k + 1+4+3+1=5k'+4$ and this cannot be divisible by 5 so the enture class of numbers x=5a+1 cannot give a result that is divisible by 5.

Now you can examine the remaining three classes of integers in the same way.
• Sep 17th 2012, 07:53 PM
MaxJasper
Re: need a easy way to solve it
$f({x})\text{=}x^5+4x^4+3x+1$

for k=1,2,3,....n

$\text{Mod}(f(k), 5) = \{4,3,2,1,1, 4,3,2,1,1, 4,3,2,1,1, ....\}$
• Sep 17th 2012, 08:18 PM
richard1234
Re: need a easy way to solve it
Just check x = 0,1,2,3,4. If x = k works, then any x that is congruent to k (mod 5) also works.