If n>1 then n^4+4 is which number?
(A) composite
(B) prime
(C) even
(D) odd
(A) composite
(B) prime
(C) even
(D) odd
n^4+4 is a composite number. If you worked on this for any length of time without success, you'll kick yourself now: n4+ 4 can be factored. Here it is:
n4 + 4
n4 + 4n2 + 4 - 4n2
( n4 + 4n2 + 4) - 4n2
((n2 + 2) + 2n)((n2 + 2) - 2n)
And since n>1 you can show that both factors are larger than one. You're done.
A similar problem was suggested to me, to prove that n4-20n2+4 is not prime. This one had me stumped for a while because I was trying this:
n4 - 20n2 + 4
n4 + 4n2 + 4 - 24n2
( n4 + 4n2 + 4) - 24n2
((n2 + 2) + sqrt(24)(n))((n2 + 2) - sqrt(24(n))
But those factors, while real, are not integers. So it doesn't help in the effort to show compositude. Sorry, I can't help making up new words. Here's the real answer:
n4 - 20n2 + 4
n4 - 4n2 + 4 - 16n2
( n4 - 4n2 + 4) - 16n2
((n2 - 2) + 4n)((n2 - 2) - 4n)
Here, the first factor is greater than one if n>0 -- that's the point where the n2 and n terms overpower the constant. The second factor is greater than one if n>4, where the n2 term overpowers the n term. So the polynomial is composite for all n>4. It's composite for n=0, 1, 2, 3, and 4, too, because its values are 4, -15, -60, -95, and -60. you worked on this for any length of time without success, you'll kick yourself now: n4+ 4 can be factored. Here it is:
n4 + 4
n4 + 4n2 + 4 - 4n2
( n4 + 4n2 + 4) - 4n2
((n2 + 2) + 2n)((n2 + 2) - 2n)
And since n>1 you can show that both factors are larger than one. You're done.
A similar problem was suggested to me, to prove that n4-20n2+4 is not prime. This one had me stumped for a while because I was trying this:
n4 - 20n2 + 4
n4 + 4n2 + 4 - 24n2
( n4 + 4n2 + 4) - 24n2
((n2 + 2) + sqrt(24)(n))((n2 + 2) - sqrt(24(n))
But those factors, while real, are not integers. So it doesn't help in the effort to show compositude. Sorry, I can't help making up new words. Here's the real answer:
n4 - 20n2 + 4
n4 - 4n2 + 4 - 16n2
( n4 - 4n2 + 4) - 16n2
((n2 - 2) + 4n)((n2 - 2) - 4n)
Here, the first factor is greater than one if n>0 -- that's the point where the n2 and n terms overpower the constant. The second factor is greater than one if n>4, where the n2 term overpowers the n term. So the polynomial is composite for all n>4. It's composite for n=0, 1, 2, 3, and 4, too, because its values are 4, -15, -60, -95, and -60.
If you worked on this for any length of time without success, you'll kick yourself now: n4+ 4 can be factored. Here it is:
n4 + 4
n4 + 4n2 + 4 - 4n2
( n4 + 4n2 + 4) - 4n2
((n2 + 2) + 2n)((n2 + 2) - 2n)
And since n>1 you can show that both factors are larger than one. You're done.
A similar problem was suggested to me, to prove that n4-20n2+4 is not prime. This one had me stumped for a while because I was trying this:
n4 - 20n2 + 4
n4 + 4n2 + 4 - 24n2
( n4 + 4n2 + 4) - 24n2
((n2 + 2) + sqrt(24)(n))((n2 + 2) - sqrt(24(n))
But those factors, while real, are not integers. So it doesn't help in the effort to show compositude. Sorry, I can't help making up new words. Here's the real answer:
n4 - 20n2 + 4
n4 - 4n2 + 4 - 16n2
( n4 - 4n2 + 4) - 16n2
((n2 - 2) + 4n)((n2 - 2) - 4n)
Here, the first factor is greater than one if n>0 -- that's the point where the n2 and n terms overpower the constant. The second factor is greater than one if n>4, where the n2 term overpowers the n term. So the polynomial is composite for all n>4. It's composite for n=0, 1, 2, 3, and 4, too, because its values are 4, -15, -60, -95, and -60. you worked on this for any length of time without success, you'll kick yourself now: n4+ 4 can be factored. Here it is:
n4 + 4
n4 + 4n2 + 4 - 4n2
( n4 + 4n2 + 4) - 4n2
((n2 + 2) + 2n)((n2 + 2) - 2n)
And since n>1 you can show that both factors are larger than one. You're done.
A similar problem was suggested to me, to prove that n4-20n2+4 is not prime. This one had me stumped for a while because I was trying this:
n4 - 20n2 + 4
n4 + 4n2 + 4 - 24n2
( n4 + 4n2 + 4) - 24n2
((n2 + 2) + sqrt(24)(n))((n2 + 2) - sqrt(24(n))
But those factors, while real, are not integers. So it doesn't help in the effort to show compositude. Sorry, I can't help making up new words. Here's the real answer:
n4 - 20n2 + 4
n4 - 4n2 + 4 - 16n2
( n4 - 4n2 + 4) - 16n2
((n2 - 2) + 4n)((n2 - 2) - 4n)
Here, the first factor is greater than one if n>0 -- that's the point where the n2 and n terms overpower the constant. The second factor is greater than one if n>4, where the n2 term overpowers the n term. So the polynomial is composite for all n>4. It's composite for n=0, 1, 2, 3, and 4, too, because its values are 4, -15, -60, -95, and -60.
