# Thread: Fraction in Lowest terms

1. ## Fraction in Lowest terms

Prove $\frac{n^3+2n}{n^4+3n^2+1}$ is a fraction in its lowest terms

Rewriting
$\frac{n(n^2+2)}{(n^2+1)^2+n^2}$

The numerator is a product, so we shall consider them separately

$\frac{n^2+2}{(n^2+1)^2+n^2}$
Let $z=n^2+1$

$\frac{z+1}{z^2+z-1}$. Numerator and denominator share no common factors

Consider
$\frac{n}{(n^2+1)^2+n^2}$
This also has no common factors

Hence
$\frac{n^3+2n}{n^4+3n^2+1}$
has no common factors
QED

Is this proof valid?

2. It believe it is correct, but whether it is acceptable to an instructor depends on your course. You may have studied an algorithm to prove similar claims, and then you need to go over the steps without justifying them. However, for this to look like a proof to an outsider, it should include not only computational steps, but also invocation of some lemmas that prove that the computations indeed show what you claim they do.

Originally Posted by I-Think
Rewriting
$\frac{n(n^2+2)}{(n^2+1)^2+n^2}$

The numerator is a product, so we shall consider them separately
Here one should say that n and n^2 + 1 are irreducible (why?) and so, if some polynomial divides n(n^2+2), then it is either n or n^2+2, using something like Euclid's lemma for polynomials. Therefore, it is sufficient to check whether the denominator is divisible by these two polynomials. Then one should say a word or two why $z+1$ does not divide $z^2+z-1$ and $n$ does not divide $(n^2+1)^2+n^2$.

3. I think that's correct. Here is another way to approach it:

Write $n^3+2n = n(n^2+2)$. Since $n^2+2$ is irriducible over $\mathbb{Q}$, either $n$ or $n^2+2$ must be a factor of
$f(n) = n^4+3n^2+1$ for the fraction not to be on the lowest form. It's obvious that $n$ isn't a factor of
$f(n)$. For $n^2+2$ to be a factor, we must have $n^4+3n^2+1 = (n^2+2)(an^2+bn+c)$, for some
$a, b, c \in\mathbb{R}$. Expanding the RHS we have $n^4+3n^2+1 = an^4+bn^3+(c+2a)n^2+2bn+2c$.
Comparing the coefficients we have (from the coefficient of [LaTeX ERROR: Convert failed] ) $a = 1$, and (from the coefficient of $n^2$)
$c+2a = 3$, so $c = 3-2a = 1$. But (from the coefficient of $n^0$) $2c = 1$, so $c = \frac{1}{2}$. A contradiction.