Wolfram gives this as solution:
solve x^5+4*x^4+3*x+1-5*k=0 for x - Wolfram|Alpha
What d'heck does it mean...Plato?
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?
(sorry for bad english:/)
Wolfram gives this as solution:
solve x^5+4*x^4+3*x+1-5*k=0 for x - Wolfram|Alpha
What d'heck does it mean...Plato?
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:
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:
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
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.