Question

Find all the solutions such that (n^4+4^n) is a prime number.

Answer 1

I don't really think you can predict a prime number...

Answer 2

Wait for this type of questions usually there are only few and then we have to prove that for the other it ain't prime

Answer 3

n=1: 5 is a prime number
n=2: 32 is not a prime number (2^5)
n=3: 145 is not a prime number (5*29)
n=4: 512 is not a prime number (2^9)
n=5: 1649 is not a prime number (17*97)

Answer 4

Hmm... how can we prove that the others are not prime number

Answer 5

Any other takers? Can you tag @kc_kennylau ?

Answer 6

@satellite73 @amistre64

Answer 7

n^4+4^n

if n is even, then there are no solutions as (2k)^4 + 4^(2k) = 2^4(k^4+4^k)
which is divisible by 16 always

Answer 8

if n is odd :

\(n^4 + 4^n \equiv 1 + (-1) \equiv 0\mod 5\)

\(\implies \) never a prime except when \(n=1\)

Answer 9

you need to consider \(n = 5k\) case separately as \((5k)^4 \not \equiv 1\mod 5\)

Answer 10

n^4 + 4^n = (n^2)^2 + (2^n)^2 + 2. n^2. 2^n - 2. n^2 .2^n

= (n^2 + 2^n)^2 - n^2 . 2^(n+1)
= (n^2 + 2^n)^2 - (n. 2^((n+1)/2))^2
= (n^2 + 2^n + n. 2^((n+1)/2))(n^2 + 2^n - n. 2^((n+1)/2))
Out required expression is prime is any of the factors is 1 resulting in other being prime.
Let us prove (n^2 + 2^n + n. 2^((n+1)/2)) is always greater than 1 for n>=1.
We shall try it by induction.
For n=1, P(n) =5. => true
let us assume P(k) is always true i.e. (k^2 + 2^k + k. 2^((k+1)/2) > 1
Now we shall check for P(k+1)
(k+1)^2 + 2^(k+1) + (k+1). 2^((k+2)/2)
=k^2 +2k +1 +2.2^k + (k+1). 2^(k+2)/2
=k^2+2^k+ k. 2^((k+1)/2) + something positive
hence p(k+1) is always true.
therefore, (n^2 + 2^n + n. 2^((n+1)/2)) is always greater than 1 for n>=1.
Now we shall check for (n^2 + 2^n - n. 2^((n+1)/2))
n=1 yields 1. Also, for n=1, (n^2 + 2^n + n. 2^((n+1)/2)) = 5, hence n=1 will yield a prime number.
Now we shall prove for n>1,n^4 + 4^n = (n^2)^2 + (2^n)^2 + 2. n^2. 2^n - 2. n^2 .2^n

= (n^2 + 2^n)^2 - n^2 . 2^(n+1)
= (n^2 + 2^n)^2 - (n. 2^((n+1)/2))^2
= (n^2 + 2^n + n. 2^((n+1)/2))(n^2 + 2^n - n. 2^((n+1)/2))
Out required expression is prime is any of the factors is 1 resulting in other being prime.
Let us prove (n^2 + 2^n + n. 2^((n+1)/2)) is always greater than 1 for n>=1.
We shall try it by induction.
For n=1, P(n) =5. => true
let us assume P(k) is always true i.e. (k^2 + 2^k + k. 2^((k+1)/2) > 1
Now we shall check for P(k+1)
(k+1)^2 + 2^(k+1) + (k+1). 2^((k+2)/2)
=k^2 +2k +1 +2.2^k + (k+1). 2^(k+2)/2
=k^2+2^k+ k. 2^((k+1)/2) + something positive
hence p(k+1) is always true.
therefore, (n^2 + 2^n + n. 2^((n+1)/2)) is always greater than 1 for n>=1.
Now we shall check for (n^2 + 2^n - n. 2^((n+1)/2))
n=1 yields 1. Also, for n=1, (n^2 + 2^n + n. 2^((n+1)/2)) = 5, hence n=1 will yield a prime number.
I am unable to prove till now that this expression is always >1 for n>1 .
Will try this later. Got to go. :P

Answer 11

The smallest acceptable positive integer value of n is 3, because n is odd.
n^4+4^n=(n^2+2^n )^2-2(n^2 )(2^n )
As we can see that n is odd, 2n+1 is even and hence we can write:
(n^2+2^n+2^((n+1)/2).n)(n^2+2^n-2^((n+1)/2).n)= p, where p is an odd prime number.
Since p is a prime, the lesser factor must equal 1 and the greater factor must equal p.
(n^2+2^n-2^((n+1)/2).n)=1
Checking a few initial values of odd n (e.g. n = 3 or 5) we observe that the above expression is greater than 1.
Let us now use Mathematical Induction to show that for odd n, the above expression > 1.
P(k): (k^2+2^k-2^((k+1)/2).k)>1 for some odd k.
P(k+2): ((k+2)^2+2^((k+2) )-2^((k+2+1)/2).(k+2) )=k^2+4k+4+2^k+3.2^k-2^((k+1)/2) k-2^((k+1)/2)-4.2^((k+1)/2)
=(k^2+2^k-2^((k+1)/2).k)+4k+4+3.2^k-2^((k+1)/2) (k+4)
>1+4k+4+3.2^k- 2^((k+1)/2) (k+4)
A quick calculation proves the above inequality for k = 3.
We now need to show that 4k+4+3.2^k-2^((k+1)/2) (k+4)> 0
Now,3.2^k=2.2^k+2^k=2^((k+1)/2).[2^((k+1)/2)+2^((k-1)/2) ]
We now observe that for k > 3, the above expression becomes greater than 2^((k+1)/2) [k+4] but we are going to prove this also.
It is analogous to showing 2^((k+1)/2)>k and 2^((k-1)/2)≥4.
Observe the latter holds for k > 1 (odd k).
We now prove the former. Let k+1 = 2m for an integer m. Then we need to show
2^m>(2m-1)
Binomially expanding,
2^m=(1+1)^m=1+m+⋯+m+1>2m+2>2m-1
Thus our proof is complete.
There is no n > 1 for which the given expression is a prime.

This is the solution I made.