Derek Sequence

Let the terms be 1, a^2 and a^4

(a^4 + a^2 + 1)(a^2-1) = a^6 - 1 = (a^3-1)(a^3+1)

Observe that a^4 + a^2 + 1 is a prime

Thus a^3 - 1 or a^3 + 1 is a multiple of a^4+a^2+1.

But for values of a > 1, a^4 + a^2 + 1 > a^3 +1

Thus the only permissible ratio is a = 1, and its easy to verify it works - 3 is a prime.

Let $x^2$ be the common ratio. The sum of the $3$ terms is then $1+x^2+x^4=(1+x^2)^2-x^2=(1+x+x^2)(1-x+x^2)$ , which is clearly composite if $1-x+x^2>1$ . This happens for every $x>1$ . It remains to check whether this sum is prime, when $x=1$ , which it is.

Given it is a geometric progression of $1 + r + r^{2} = p$ , where $p$ is a prime number. Since $r = 1^{2}, 2^{2}, ..., 1000^{2}$ , therefore $1≤ \sqrt{r} ≤1000$ .

From the equation $1 + r + r^{2} = p$ , completing the square will result in $(r+1)^{2} - r = p$ . Then, further factorization obtains the equation $(r +1 - \sqrt{r})(r + 1 + \sqrt{r}$ ) = p)

Taking the factor $(r + 1 - \sqrt{r})$ , the solution results in $r = \sqrt{r}$ . Therefore, this equation only holds when $r = 1$ . Number of $r$ -values is $1$ , therefore answer is $\boxed{1}$ .

the terms would be of the form $1, r^{2}, r^{4}$

$1 + r^2 + r^4 = (r^2 + 1 + r)(r^2 + r - 1)$ must be a prime.

so one of the factors must be $1$ .

equation both the factors to $1$ , we see that $r = 1$ gives us the solution we seek.

the sum then being $3$ which is a prime.

Note that the sequences will be in the form: 1, x^2, x^4

The the sum can be factorized: 1 + x^2 + x^4 = (1 - x + x^2)(1 + x + x^2)

It will be a prime number when the smallest factor be equal to 1:

1 - x + x^2 = 1 Since x > 0 we get that x =1

1 + x^2 + x^4 = 3 is the only solution

Write the equation given on the problem. Note that the first term is 1 and the common ratio is the square of an integer between 1 and 1000. So, the equation is: 1 + x^2 + x^4 (as prime).

= x^4 + 2x^2 + 1 - x^2 = (x^2 + 1)^2 - x^2 = (x^2 + 1 + x)(x^2 +1 - x).

Substituting as x = 1 gives (1 + 1 + 1)(1) = 3. Hence, a prime number.

For any other values of x, the value of each bracket greater than 1 and 1 + x^2 + x^4 is the product of at least two factors.

Thus, 1 + 1^2 + 1^4 is the only value for x. (x = 1)

We can represent the 3 terms in the form: $1, q^2, q^4$ . Their sum is: $S = 1 + q^2 + q^4 = (1 + q^2)^2 - q^2 = (1 + q +q^2)*(1 - q + q^2)$ . For $S$ to be prime, $1 - q + q^2$ must equal $1$ . That is only true for $q = 1$ , hence there is only one geometric sequence that satisfies the aforementioned.

Because we are dealing with geometric sequences whose common ratio is the square of an integer, and we're given our first term is 1, the sum we are looking at is $1+n^2+n^4$ , for some integer n. Now note that $(n^2+n+1)(n^2-n+1)=n^4+n^2+1$ , so we must find some n such that $n^2+n+1=1$ or $n^2-n+1=1$ . The resulting two equations are $n^2 \pm n=0$ , which we see have three integer solutions, $n= \pm 1$ and $n=0$ . Because the common ratio is the square of an integer between 1 and 1000, we can disregard $n=0,-1$ , and our only solution is when the common ratio is 1.

Let the ratio be $r$ . The sum is $1 + r^2 + r^4$ . Manipulation: $1 + r^2 + r^4 = r^4 + 2 r^2 + 1 - r^2 = (r^2 + 1)^2 - r^2 = (r^2 + 1 - r)(r^2 + 1 + r)$ (I believe this is an identity named after Sophie Germain.)

Both quadratics are positive and increasing for integers $r \geq 1$ . Furthermore, the first is always $\geq 3$ on the same set, so the second must be $1$ and the first prime for the product to be prime. When $r = 1$ , the first is $3$ and the second is $1$ . When $r = 2$ , the first is $7$ and the second is $3$ . This already means that all products with $r > 1$ will have two factors above $1$ , i.e. be composite. Therefore, only the first number works.

The answer is $\boxed{1}$

Let the 3-term geometric sequence be $1, n^{2}$ and $n^{4}$ , where $n^{2}$ is the common ratio and $n$ is an integer between 1 and 1000 inclusively. Thus, $1+n^{2}+n^{4}=k$ , where $k$ is a prime number.

$1+n^{2}+n^{4}=1+2n^{2}+n^{4}-n^{2}$

$=(n^{2}+1)^{2}-n^{2}$

$=(n^{2}+n+1)(n^{2}-n+1)$

Thus, in order for $k=(n^{2}+n+1)(n^{2}-n+1)$ , one of these two factors must equal to one.

If $n^{2}+n+1=1$ , then $n^{2}+n=0$ and $n=0$ or $n=-1$ . However, these are inadmissible answers, as $n$ must be between 1 and 1000.

If $n^{2}-n+1=1$ , then $n^{2}-n=0$ and $n=0$ or $n=1$ . Here, we see that $n=1$ is the only possible solution.

Thus, there is only one solution (when the common ratio is $1$ and the sum of the 3 terms is $3$ ).

The sequence is, naturally, given as:

${1, k, k^2} \forall k \in Z, 1 \leq k \leq 1000$

The sum is, then, equal to:

$k^4 + k^2 + 1 = (k^2 + 1)^2 - k^2 = (k^2 + 1 + k)(k^2 + 1 - k)$

Now, this sum will only be prime if either one of these factors equals $1$ , and the other is a prime number. For the first condition, we see that

$k = 0$ and $k = 1$

are the only solutions. Since $k \neq 0$ , because the common ratio $k$ has to be between $1$ and $1000$ inclusive, $k = 1$ . Checking for the second condition,

$k^2 + k + 1|_{k = 1} = 3$

which is a prime.

Hence, there is only $\boxed {1}$ solution.

Let our chosen ratio be expressed as $r$ . Then the sum of the series is $r^4+r^2+1$ . Noticing the all even powers and the prime condition, we are motivated to consider a factorization as a difference of squares. It catches our eye that $(r^2+1)^2$ will be just larger than the sum. So, we write $r^4+r^2+1=(r^2+1)^2-(r)^2=(r^2+r+1)(r^2-r+1)$ For this to be prime, we require one of the factors to be equal to $\pm 1$ , and obviously $r^2+r>0$ , so $r^2-r+1=1\implies r^2=r\implies r=1$ Thus, there is exactly $\boxed{1}$ solution.

The geometric sequence would look like $1 , r^{2} , r^{4}$ . We add these terms to get $r^{4} + r^{2} + 1$ . We can rewrite the expression to be $r^{4} + 2r^{2} + 1 - r^{2}$ . This expression can be simplified to $(r^{2} + 1)^{2} - r^{2}$ .

Going even further, we can say $(r^{2} + 1 + r)(r^{2} + 1 - r)$ . This means that $r^{4} + r^{2} + 1$ is going to be a product of two integers. We can see that $r^{4} + r^{2} + 1$ will never be a prime number except for when $r = 1$ .

The sum is of the form: 1 + r^2 + r^4 = (1 + r^2)^2 - (r)^2

As it is a difference of squares: a^2 - b^2 = (a+b)(a-b) .... where a and b are natural numbers,

this can be prime, if and only if: (a-b) is equal to 1; and (a+b) is prime

thus, 1+ r^2 - r = 1 giving r = 1 for which 1+ r^2 + r = 3, is prime ... thus only possible for r = 1

Let's call our integer k; 1<=k<=1000. The formula for a geometric series is S = a(1-r^n)/(1-r), where a is initial term, r is the ratio and n is number of terms. Now we know a=1, r=k^2, and n=3 so we plug these in to get (1-k^6)/(1-k^2) = (k^6-1)/(k^2-1). We wish to find all k such that this expression is prime.

We know this should divide to be an integer somehow; let's try factoring. Using difference of squares gives (k^3+1)(k^3-1)/(k+1)(k-1). Applying sum/difference of cubes to the top reduces it to (k+1)(k^2-k+1)(k-1)(k^2+k+1)/(k+1)(k-1).

If k=1, the expression is 3, which is indeed prime. If k isn't 1, we divide out the common factors on top and bottom (since neither is 0) to get (k^2-k+1)(k^2+k+1) which must be prime - so one of these factors is 1 and the other is the prime itself. (k^2-k+1) is the smaller number - k is positive - so it must equal 1. But solving for k again we have k=1 and S=3. Thus k=1 is the ONLY solution.