A couple of not so simple problems from old contests

[myro111]

I found some problems from old contests in my book and I need some help solving them since I could not find the solutions online.

Some USA contest problems:

1.(1977.)Let a and b be 2 solutions of $$x^4+x^3-1=0$$ .Prove that $$a*b$$ is the solution of $$x^6+x^4+x^3-x^2-1=0$$

2.(1983.).Prove that all the solutions of $$x^5+ax^4+bx^3+cx^2+dx+e=0$$ are real if $$2a^2<5b$$

German contest

3.(1977.)How many pairs of numbers p,and q from $$N$$ which are smaller than 100 and for which $$x^5+px+q=0$$ has a rational solution exist ?

Moscow olympiad

4.(1951.) Dividing the polynomial $$x^1^9^5^1-1$$ with $$P(x)=x^4+x^3+2*x^2+x+1$$ we get a quotient and remainder.What is the coefficient next to $$x^1^4$$ in the quotient?

5.(1955.)If $$p/q$$ is the root of the polynomial $$f(x)=a[0]*x^n+a[1]*x^n^-^1+...+a[n]$$ and p and q don't have common divisors.If $$f(x)$$ has integer coefficients then prove that $$p-k*q$$ is a divisor of $$f(k)$$ for every integer k.

Thank you very much!

 

[verty]

What have you tried so far? Do you have ideas how to tackle them?

 

[myro111]

1. Divide the second polynomial with x-ab and I get as a remainder $$(ab)^6+(ab)^4+(ab)^3-(ab)^2-1$$ The remainder is the second polynomial but instead of x it is ab.I got this idea because if we have a polynomial P(x),a is the root only and only if P(x) is divisible by x-a,and that fact is well known.The other idea was Vietes formulas but couldn't get anything.

2.No idea.

3.No idea

4.No idea.

5.No idea.

 

[verty]

I'm kidding myself that I have anything close to the knowledge required to solve these, but...

Write x^6 + x^4 + x^3 - x^2 - 1 = 0 as x^4 + x^3 - 1 = x^2 (1 - x^4), substitute x = a, that should help you to get a range of values for ab.

 

[myro111]

Dont worry I solved it another way.But thanks for the try.I used Vietes formulas