Primes N^4 + 4

By the Sophie Germain identity, $N^4+4=(N^2+2N+2)(N^2-2N+2)$ Since $N^2+2N+2=(N+1)^2+1$ and $N^2-2N+2=(N-1)^2+1$ , both factors are nonnegative, so if they multiply to a prime, one must be 1.

If $N^2+2N+2=1$ , $N^2+2N+1=0$ , and $N=-1$ , but $N$ is positive.

If $N^2-2N+2=1$ , $N^2-2N+1=0$ , and $N=1$ , which is a solution.

Indeed, $1^4+4=5$ , which is a prime.

Therefore, there is $\boxed{1}$ solution for $N$ .

Motivation: Whenever the problem discusses primes and I see fourth powers, the Sophie Germain identity is a good thing to try.

We know that $a^2+b^2=(a+b)^2-2ab$ . Using this $n^4+2^2=(N^2+2)^2-4N^2=(N^2+2N+2)(N^2-2N+2)$ . So if it is a prime we must have $N^2-2N+2=1\Rightarrow (N-1)^2=0$ . So $N=1$ and by checking we find this is correct indeed.

$N^4 + 4 = (N^2 + 2)^2 - 4N^2 = (N^2 + 2)^2 - (2N)^2 =$

$(N^2 + 2N + 2)(N^2 - 2N + 2)$

$\text{because } a^2 + b^2 = (a+b)^2 - 2ab \text{ and } a^2 - b^2 = (a+b)(a-b)$

$\text{Since }N^4 + 4 \text{ is prime and }N \text{ is a positive integer, }$

$N^2 - 2N + 2 = 1$

$N^2 - 2N + 1 = 0$

$(N-1)^2 = 0$

$N = 1$

$\text{Therefore there is only 1 possible }N \text{that fulfills the condition.}$

N^4+4 = (N^2+2+2N)(N^2-2+2N) If the desired expression is prime, then one of the factors must be equal to one, which only occurs when N=1.

(n square+2n+2)(n square-2n+2)=N power 4 it can only be prime if either is 1. on solving we get n=+1 & -1 as n is prime n=1 only only solution

[My apologies if the formatting's not great; first time using LaTeX.]

There is only one positive integer satisfying the given condition, namely N=1.

We note that

$N^4 + 4 \\ = (N^4 + 4N^2 + 4) - 4N^2 \\ = (N^2 + 2)^2 - (2N)^2$

which is a difference of squares and so can be factored into

$(N^2 + 2N +2) (N^2 -2N + 2)$ .

Thus $N^4 + 4$ is only prime if one of the above factors equals 1. Since N is a positive integer, the first factor above is clearly $\geq$ 5, so this only leaves the possibility that the second above factor equals 1. Then

$N^2 -2N + 2 = 1 \\ N^2 - 2N + 1 = 0 \\ (N - 1)^2 = 0$

The only solution to the above, N=1, clearly makes $N^4 + 4$ prime, so it is the only solution to the given problem.

My first thought was to try congruence which did not help me much. Next, I tried factoring. This one did the job for me.

This is how I did it :

$N^4+4$ $\Rightarrow N^4 + 4N^2 + 4 - 4N^2$ $\Rightarrow (N^2 + 2)^2 - (2N)^2$ $\Rightarrow (N^2 + 2 + 2N)(N^2 + 2 - 2N)$

So, the given expression is a prime only if one of the factors is 1. Since N is a positive integer, $N^2 + 2 - 2N = 1$ $\Rightarrow (N-1)^2 = 0$ $\Rightarrow N = 1$ So we conclude, there is only 1 value of N satisfying all conditions.

We know that $N^4+4=(N^2+2N+2)(N^2-2N+2)$ So, for $N^4+4$ to be a prime number, either: $N^2+2N+2=1$ or $N^2-2N+2=1$ Solving, we get $(N+1)^2=0$ and $(N-1)^2-0$ We know that $N=1$ . (Ignoring negative since N is positive) Plugging the roots into $N^4+4$ , we get 5 for each of them. Therefore, there is only $\boxed{1}$ positive integer N.

We can check pretty quickly that there is only 1 prime in the form of $N^4+4$ . Calculating the first few positive odd integers, (we ignore the evens since we know that even+even=even) we can clearly see that the last digit of the end result is always 5 (making it a multiple of 5), except for $N$ with 5 as their last digit. However, $N=5$ doesn't work, and any number after that is divisible by 5. Therefore, seeing that only 1 positive integer N works (which ironically is when N=1).

Only for 1 the N^4 + 4 is prime and for no more numbers the sum is prime thats y.

$n^{4} + 4 = (n^{2} - 2n + 2) \times (n^{2} + 2n + 2)$

Now $(n^{2} + 2n + 2) >= 5$ and for $n = 1$ we have equality. The only chance to have prime number is this term to be prime (5 is prime fortunately) and $(n^{2} - 2n + 2) = 1$ which is equivalent to $(n-1)^{2} = 0$ thus only $n = 1$ is solution.

The prime number must be greater than 4. So, it is obviously odd. First we check for the last digit of the resulting number. You will find that $1^4, 3^4, 5^4, 7^4, 9^4$ all end with 1. So the resulting number has last digit 5. If the number has two or more digits, it will have 5 as its factor and hence, won't be prime. So, the $only$ prime of the given form is $5$ .

We can factor $N^4+4$ as $(N^2+2N+2)(N^2-2N+2)$ If $N=1$ , we find that $N^2+2N+2=5$ while $N^2-2N+2=1$ Since $5\times1=5$ and 5 is prime, $N=1$ works.

However, if we go beyond $N=1$ , we can see that both terms in the factorization will be a positive integer greater than 1. This means that $N^4+4$ would have to be composite.

Therefore, there is only $\fbox{1}$ positive integer N in which $N^4+4$ is prime. $\ \blacksquare$

We must show that $l=N^4+4$ is prime only for a finite number of positive integers, that is $N>0$ .

Note that we have a suggestive factoring: $l=N^4+4N^2+4-4N^2=(N^2+2)^2-4N^2$

Further factoring by difference of squares yields: $(N^2+2)^2-4N^2=(N^2+2-2N)(N^2+2+2N)=l$

Clearly, then, the only time that $l$ is prime is when one of $(N^2+2-2N),(N^2+2+2N)$ is a unit. That happens only when $N=1$ for $N>0$ , hence, we are done and $l=5$ is our only prime number.

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Coming back from a while ago, for some reason Brilliant marked this as not being solved. Remembering less number theory (and especially not remembering Sophie Germain's identity) than I did before, I solved this a different way my second time.

I plugged in three odd numbers $1, 3, 5$ , since plugging in even numbers yields an even solution, and quickly realized that all of them were multiplies of five. Of course, there are only 4 numbers co-prime to 5, so we can test all of them to show that $N^4 \equiv 1 \mod 5$ for any $N$ . Doing this calculation for $1, 2$ yields the entire set of results (as $1 \equiv -4$ such that after squaring, they're the same value), confirming the (now stronger!) statement that, in fact, every number of the form $N^4+4$ is divisible by either 4 or 5 with $N=1$ giving the only prime result (namely, 5).

n=1 only works lol

By the Sophie Germain Identity , the expression can be factored into $(N^2+2-2N)(N^2+2-2N)$ .

If an integer is prime, then one of its factors must be one. With this statement, we arrive at two cases.

Case $1$ : $N^2+2-2N$ is $1$ .

$N^2-2N+1=0$

$N_{1}=-1, N_{2}=1$

Case $2$ : $N^2+2+2N$ is $1$ .

$N^2+2N+1=0$

$N_{3}=-1, N_{4}=-1$

From all possible values of $N$ , only one meets the criteria that $N$ is a positive integer. So, the answer is $\boxed{1}$ .

To determine the number of possible positive integer values of $N$ such that $N^4 + 4$ is a prime, we must first factor $N^4 + 4$ . But how can we do this? We cannot factor $N^4 + 4$ directly but we can introduce another term in order for it to be factored specifically the term $4N^2$

$N^4 + 4 = N^4 + 4 + 4N^2 - 4N^2$

Rearranging the terms and using factoring a perfect square trinomial and a difference of two squares yield,

$N^4 + 4 = N^4 + 4N^2 + 4 - 4N^2$

$N^4 + 4 = (N^2 + 2)^2 - 4N^2$

$N^4 + 4 = (N^2 + 2 + 2N)(N^2 + 2 - 2N)$

By using the definition of a prime number which is "the factors of a prime number is 1 and itself",

$N^2 + 2 + 2N = 1$ or $N^2 + 2 - 2N = 1$

Case 1:

$N^2 + 2 + 2N = 1$

$N^2 + 2N + 1 = 0$

$(N + 1)^2 = 0$

$N = -1$

But N must be a positive integer, therefore there are no possible solutions.

Case 2:

$N^2 + 2 - 2N = 1$

$N^2 - 2N + 1 = 0$

$(N - 1)^2 = 0$

$N = 1$

1 is a positive integer therefore $N = 1$ . Since 1 is the only possible value of N, the number of possible positive integer values of $N$ such that $N^4 + 4$ is a prime is 1

We factor.

$N^4+4 = (N^2)^2+2^2 = (N^2+2)^2-2(N^2\cdot 2) = (N^2+2)^2 - (2N)^2 = (N^2-2N+2)(N^2+2N+2) = p$ where p is a prime. Notice that one factor must be equal to 1 to yield a prime. This gives $N = -1, 1$ respectively. However, N must be positive so we get only $\boxed{1}$ solution, which is N=1, yielding 5.

We can use the Sophie Germain identity...

$a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)$

Substituting $a=N$ and $b=1$ , we get...

$N^4 + 4 = (N^2 + 2N + 2)(N^2 - 2N + 2)$

We need one of the factors to be $1$ ...

Since, $N > 0$ , so $N^2 + 2N + 2 > 1$ ...

Hence, if $N^2 - 2N + 2 = 1 \Longrightarrow N^2 - 2N + 1 = 0 \Longrightarrow (N-1)^2 = 0 \Longrightarrow N = 1$

If $N = 1$ , then $N^4 + 4 = 5$

Hence, there's only $\boxed{1}$ such $N$ ...

By Fermat's Little Theorem, for any integer a, a^4 = 1 mod 5. Now, if we add 4, the result is a multiple of 5. The only multiple of 5 that is prime is.... 5 itself. Checking the smallest positive integer, we find that 1 produces the prime 5; hence it is the only solution.

When n is 1, we gwt 5 which is peime. Suppose that n is at least 2. If n is even, the sum of them is even. If n is odd, the sum doesn't congruent to 0 mod 5. Thus, n equals to 1 is the only solution.

I solved it in a much more simple way, lol

since

• 0 raised to any power (except zero of course) will be zero
• 1 raised to any power is 1
• 2 raised to any power(except zero) the last digit will always be 2, 4, 8 or 6
• 3 raised to any power the last digit will always be 3, 9,7 or 1 & so on till we reach 9

therefore, the only possible solution ( as the last digit of any $N^{4} + 4$ will be either even or 5, and at both possibilities the value will not be prime, Except when 5 is the only digit in the number) is 1

Please tell me if my solution has any flaws. :)

N^4 for digits ending with 1 ends with 1 digits ending with 2 ends with 6 digits ending with 3 ends with 1 digits ending with 4 ends with 6 digits ending with 5 ends with 5 digits ending with 6 ends with 6 digits ending with 7 ends with 1 digits ending with 8 ends with 6 digits ending with 9 ends with 1 digits ending with 0 ends with 0

on adding 4, we get 4, 5, 9, 0 in units place. 0 and 4 makes number even. on subtracting 4 from prime numbers ending with 9, we do not get number in form N^4. digits ending with 5 have factor 5.

Hence only prime number is 5.

1^4=1 2^4 =4*4=6(unit digit only) 3^4=`1.....5^4=5, if a even number (4) added with odd number it will become prime.so 2^4,4^4,6^4,8^4,...etc has even unit digit .remaining we have 1,3,5,7,9.3^4+4=85,5^4+4=629/17=4,7^4+4=5(unit digit),9^4+4=5. finally 1^4+4=5

$n^4+4=(n^2+2n+2)(n^2-2n+2)$ therefore, $n^2+2n+2=1$, like this n=1