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I found this function, and i have problems with the demonstration of truth or falseness of this afirmation, so some one can help me?

Let S and P be the sum and the product of the divisors of a number T $\in$ $\mathbb Z$ and T different of 1.In the case T is a prime number, the roots are +i, -i. In this last case a think the function prove what a prime number don't have more divisors.

I want to prove when T has an four dividers the roots of the f(x) is the non trivial divisors of T.

f(X) = T$x^2$ + $T^2$x + Tx - STx + P, dividing by T we have:

f(x) = $x^2$ + (T + 1 - S)x + $\frac{P}{T}$

But P is equal at the product of divisors of T, when the divisors of T is even P = $T^\frac{n}{2}$, n is the amount of divisors of T (we call $\frac{n}{2}$ = k), when the amount of divisors is even, so $\frac{P}{T}$ = $\frac{T^\frac{n}{2}}{T}$ = $\frac{T^k}{T}$ = $T^{k-1}$, so

f(x) = $x^2$ + (T + 1 - S)x + $T^{k-1}$

I tried prove it some times, but i have problems with the arguments, so some one can give me a hint already help. itried what the product of the roots resolve the problem, i think is correct of this form.

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  • $\begingroup$ I recommend you check carefully what you have written and try to explain it clearly, step by step. What do you mean when you say $S$ and $P$ are the sum and product of $T$? Where did the number $n$ come from? $\endgroup$ – David K Mar 8 at 18:14
  • $\begingroup$ Also, it is good you used some MathJax, but it is much better to put an entire formula into MathJax all at once, not just bits and pieces like when you want an exponent or a particular kind of arrow. $\endgroup$ – David K Mar 8 at 18:16
  • $\begingroup$ Also, the claim $f(x) = Tx^2 + T^2x + Tx - STx + P \Rightarrow f(x) = x^2 + (T + 1 - S)x + \frac{P}{T}$ is obviously false unless $T=1.$ I have a hunch that what you wrote is very different from what you wanted to say. $\endgroup$ – David K Mar 8 at 18:19
  • $\begingroup$ Thanks David K, for the hints, i think what i arranged the issue. $\endgroup$ – Latith Mar 9 at 10:24
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Setting $f(x) = Tx^2 + T^2x + Tx - STx + P$, the roots of $f(x)$ are the values of $x$ that solve the equation $$ Tx^2 + T^2x + Tx - STx + P = 0. \tag1$$ The solutions for $x$ in Equation $(1)$ are also called the roots of the equation.

As you have shown, if $T$ has $n$ divisors then the roots of Equation $(1)$ are the same as the roots of $$ x^2 + (T + 1 - S)x + T^{(n/2)-1} = 0 . \tag2$$ (Note that if you view the polynomials as functions of $x,$ then $x^2 + (T + 1 - S)x + T^{(n/2)-1} = \frac1T f(x)$, which is not the same as $f(x)$ according to the way you defined $f(x).$ The only thing that will stay the same when you divide it by $T$ is zero. I think it's clearer to write "$=0$" when we need something to be zero.)

You claim that if $n = 4$ then the roots of Equation $(2)$ are the non-trivial divisors of $T.$ Note that if $n = 4$ then $\frac n2 - 1 = 1,$ so Equation $(2)$ simplifies to $$ x^2 + (T + 1 - S)x + T = 0 . $$

Let's try it.

For $T = 10,$ the divisors are $1,2,5,10,$ so $S = 18$ and $x^2 + (T + 1 - S)x + T = x^2 - 7x + 10.$ The roots of $x^2 -7x + 10 = 0$ are $2$ and $5,$ as predicted.

For $T = 8,$ the divisors are $1,2,4,8,$ so $S = 15$ and $x^2 + (T + 1 - S)x + T = x^2 - 6x + 8.$ The roots of $x^2 - 6x + 8 = 0$ are $2$ and $4,$ as predicted.

So that's one case where $T$ is the product of two distinct primes, and one case where $T$ is the cube of a prime. For an actual proof, you need to show that the claim is true for the product of any two distinct primes and for the cube of any prime. You also need to show that these are the only cases of numbers that can have four divisors; you might consider how we can count the divisors of a number using its prime factorization.

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