Show that $n^4-20n^2+4$ is composite when $n$ is any integer.

Question

Crux Mathematicorum June 1978,

Show that $n^4-20n^2+4$ is composite when $n$ is any integer.

My question: What is meant by 'composite'?

Besides, I found out that this expression could be factorised into $(n^2-4n-2)(n^2+4n-2)$. Is this related to the question?

Hope someone can clarify what is meant by composite and what the question actually and solve this. Thanks in advance.

Your factorisation is very useful. If you can show that both of those terms aren't 1 (or -1) for some $n$, then you have shown that for that $ n$, $n^4 - 20n^2 + 4$ can be factorised. — B. Mehta, May 20 '17 at 13:45
You've already solved the question by factorising the expression :) — Hypergeometricx, May 20 '17 at 13:50
According to @Frpzzd, composite is not a prime, how to continue from my expression to show that it's not a prime? — Mathxx, May 20 '17 at 13:51
You already have two factors so if neither of them are of absolute value $1$ or $0$ (which can be shown easily) then it is composite. — Hypergeometricx, May 20 '17 at 13:56

score 4 · Accepted Answer · edited May 22 '17 at 18:00

Given the expression $n^4-20n^2+4$

You write

"I found out that this expression could be factorised into".. $$(n^2−4n−2)(n^2+4n−2).$$ Is this related to the question?"

YES!

If neither one nor the other of the factors you found evaluates to $1, 0, -1,$ under any integer value of $n$, you will have shown that the expression can be factored into two non-zero, non-unit factors, and hence is composite.

Suppose we first try to solve whether either factor might evaluate to zero given some integer n:

$(1)\;\;n^2-4n - 2 = 0$

$(2)\;\; n^2 + 4n - 2 = 0$

By using the quadratic equation, you'll find that there are no integer roots for either. So neither factor is zero.

Similarly, you can use the quadratic equation to test out whether either factor can equal $\pm 1$

$(3)\;\;n^2-4n-2 = 1 \iff n^2-4n -3 = 0$

$(4)\;\;n^2 - 4n-2 = -1 \iff n^2 - 4n -1 = 0$.

Similarly, we see there are no integer roots $n$ when solving for $n$

$(5)\;\; n^2+4n-3 = 0, \;\;\;(6)\;\;n^2 + 4n -1 = 0$

Now, if we use common definition of a composite number as being strictly a positive integer $k$ with at least three factors (or at least one factor other than $1$ and itself $(k\geq 2)$, then the statement is not true.

Counter example under this stricter definition:

When $n=3$, the expression evaluates to: $3^4-20\times 3^2+4 = 81-180 + 4 = -95$. This is not a composite number, strictly defined, because it is not a positive integer. However, since $-95 = (-1)(1)(5)(19)$, for the purposes of this assignment, we can see that $-95$ is a negative composite number.

They do evaluate to $1$ or $-1$ for some $n$... however, we only want to consider integer values of $n$. — Franklin Pezzuti Dyer, May 20 '17 at 13:55
That's exactly what I mean, $n$ must be an integer, and hence neither of the factors evaluate to 1 or -1 when n is an integer. — amWhy, May 20 '17 at 13:59

Hypergeometricx · Answer 2 · 2017-05-20T17:04:33.327

3

Note that $$n^2-4n-2=(n-2)^2-6$$ and $$n^2+4n-2=(n+2)^2-6$$ For integer $n$, neither of $(n-2)^2$ or $(n+2)^2$ will give a value of $5,6,7$ (neither of these are perfect squares) hence neither of the factors are $-1,0,1$.

As such, the product of both factors is a composite number.

edited May 20 '17 at 17:04

answered May 20 '17 at 14:00

Hypergeometricx

22,657

Show that $n^4-20n^2+4$ is composite when $n$ is any integer.

2 Answers2