Joe's primes

We notice that $x^3+y^3 = (x+y)^3-3(x+y)(xy)$ .

Therefore we substitute into $a$ and $b$ to get $a^3-3ab=p$ for some prime $p$ By the rational root theorem, since $a$ and $b$ are integers, $a$ must be $1$ or the prime $p$ .

Checking $a=1$ we get $1-3b=p$ which does not work because $p$ and $b$ must be positive by the given conditions on the problem.

Therefore $a = p$ and we get the equation $p^3-3pb-p=0$ or $p^2-3b-1=0$ which means $p^2=3b+1$ . Looking at the squares in mod $3$ we get that $p$ must be congruent to $1$ or $2$ mod $3$ .

The restriction on $b$ is bounded from $1$ to $100$ so we see the max value $a$ can be is $17$ . Listing all the primes congruent to $1$ or $2$ mod $3$ and less than or equal to $17$ we get $2,5,7,11,13,17$ for a sum of $55$

We notice that $x^3$ + $y^3$ = $(x + y)^3$ - $3xy(x+y)$

Putting the values of $a$ and $b$ we have $x^3$ + $y^3$ = $a(a^2-3b)$

Now for this to be prime one of the factors must be 1 and the other must be a prime number.

Case 1:

$a$ = 1 and $1-3b$ is prime

this is not possible for positive values of $b$

Case 2: $a^2 - 3b$ = 1 and a is prime

i.e $\sqrt{3b +1}$ is prime .

the possible values of $a$ are $2 , 5, 7, 11, 13, 17$ which have a sum of $55.$

In the problem the value of (x^3) + (y^3) is to be a prime. So first of all the value of ( x^3) + (y^3) is to be obtained from the given values. (x^3) + (y^3) = (x+y) ( (x^2)- xy + (y^2) )..= (a)x ( (a^2)- 3b )............(A) So, (a) x ( (a^2) - 3b ) is to be a positive prime. Note that a prime can be factored into 2 ways: (1 x the prime) or (prime x 1). So (a) x ( (a^2) -3b ) can be factored into 2 ways.

Case I:..............a=1 and (a^2) - 3b = p (the prime)......so 3b = (a^2) - p = 1-p.........Since b>=1........this can never be unless p<0........so this case can't be considered...........

Case II:......a=p ..........(a^2) - 3b =1.........(B).........note that from the second equation............ 3b = (a^2)-1 = (a+1) x (a-1)........................since b is an integer.........3 must divide R.H.S.........note that gcd( (a+1, a-1) ) = +-1 or +-2........so 3 has to to divide either a+1 or a-1 but not both.

First let 3 divide a+1......so a has to be of the form 3k+2 for k=0,1,2,3...................... Plugging in k=0,1,2,3...........the diff values of a are obtained. Putting this values in (B) ..,,,the permissible values of a for which b<100 are : 2, 5, 11 , 17

Second, let 3 divide a-1............so a will be of the form a=3k+1 for k=0,1,2.... and repeating the same way as above the permissible values of a are : 7, 13

So the sum of different values of a is 2 + 5 + 7 + 11 + 13 +17 =55

$\displaystyle x^3+y^3=(x+y)^3-3xy(x+y)\Rightarrow x^3+y^3=a^3-3ab$ $=a(a^2-3b)$ Since $x^3+y^3=p$ for some prime $p$ , $a$ or $a^2-3b$ have to be $1$ . Checking when $a=1$ , we get $1-3b=p$ , which is not possible because of the conditions specified. So $a^2-3b=1$ which is equivalent to $a^2\equiv 1\pmod3$ . In addition, $x^3+y^3=a(1)=a$ . By Fermat's Little Theorem, $a^2\equiv 1\pmod3$ is always true when $a\neq3k$ for any positive integer $k$ . So this problem just reduces down to finding all primes with $a$ and $b$ less than 100. Clearly $a$ grows faster than $b$ so we need to apply the restriction on $b$ . With $b=100$ , we get $a^2-300=1\Rightarrow a<=\sqrt{301}\Rightarrow a<=17$ . Adding all the primes equal to or less than 17 that are not divisible by 3 we get $2+5+7+11+13+17=55$

In the problem the value of (x^3) + (y^3) is to be a prime. So first of all the value of ( x^3) + (y^3) is to be obtained from the given values.
(x^3) + (y^3) = (x+y) ( (x^2)- xy + (y^2) )..= (a)x ( (a^2)- 3b )............(A) So, (a) x ( (a^2) - 3b ) is to be a positive prime. Note that a prime can be factored into 2 ways: (1 x the prime) or (prime x 1). So (a) x ( (a^2) -3b ) can be factored into 2 ways.

Case I:..............a=1 and (a^2) - 3b = p (the prime)......so 3b = (a^2) - p = 1-p.........Since b>=1........this can never be unless p<0........so this case can't be considered...........

Case II:......a=p ..........(a^2) - 3b =1.........(B).........note that from the second equation............ 3b = (a^2)-1 = (a+1) x (a-1)........................since b is an integer.........3 must divide R.H.S.........note that gcd( (a+1, a-1) ) = +-1 or +-2........so 3 has to to divide either a+1 or a-1 but not both.

First let 3 divide a+1......so a has to be of the form 3k+2 for k=0,1,2,3...................... Plugging in k=0,1,2,3...........the diff values of a are obtained. Putting this values in (B) ..,,,the permissible values of a for which b<100 are : 2, 5, 11 , 17

Second, let 3 divide a-1............so a will be of the form a=3k+1 for k=0,1,2.... and repeating the same way as above the permissible values of a are : 7, 13

So the sum of different values of a is 2 + 5 + 7 + 11 + 13 +17 =55

x^3+y^3= (x+y)(x^2+y^2-xy)=a (a^2-3b)=P For P to be a prime a must be a prime and simultaneously a^2-3b must be equal to one. Putting the values of a starting from 2 to 17 (only prime numbers with both 2 and 17 inclusive). We will stop at 17 because after that b becomes greater than 100. At last add the solutions of a you get which will be equal to 55

$x^3+y^3 = (x+y)(x^2-xy+y^2) = a (a^2 - 3b)$

If this is a positive prime, then either $a = 1$ or $a^2 - 3b = 1$ . If $a = 1$ , $a^2 - 3b < 0$ . So we must have the condition that $a^2 - 3b = 1$ . As long as $a^2 < 300$ , $a$ is not divisible by 3 and $a$ is prime, it will satisfy the criteria in the problem. That leaves $a = 2, 5, 7, 11, 13, 17$ .

We have: $x^3+y^3=(x+y)(x^2+y^2-xy)=a(a^2-3b)$ . Since $x^3+y^3$ is a positive prime number, we have two case:

Case 1: a=1 and $a^2-3b$ is a positive prime number. This is impossible since $a^2-3b=1-3b<0$ when $b \geq 1$ .

Case 2: $a^2-3b=1$ and a is a positive prime number. Because $b \leq 100$ , so $a^2 \leq 301$ or $a \leq 17$ . Also, $a^2 \equiv 1$ (mod 3). Therefore, possible values of a are: 2,5,7,11,13,17.

The answer is: 2+5+7+11+13+17=55

$x^3 + y^3 = (x + y)^3 -3xy(x+y) = a^3 - 3ab = a(a^2 - 3b)$

Now, $a(a^2 - 3b)$ is a prime number so, we have $2$ case :

Case $1$ : $a = 1$ and $(a^2 -3b)$ is a prime number . But it's impossible for $b$ .

Case $2$ : $a$ is a prime number and $(a^2 - 3b) = 1$

$\Rightarrow a^2 = 1+3b$

$\Rightarrow a = \sqrt{1+3b}$

So, $\sqrt{1+3b}$ is a prime number.

Now, the possible values of $a$ are $2,5,7,11,13,17$ which have a sum of $\boxed {55}$ .

(x+y)^3 = x^3 + y^3 + 3xy(x+y) So: a^3 = x^3 + y^3 +3ba x^3 + y^3 = a(a^2 - 3b) (x^3 + y^3) is a prime number, so [a(a^2 - 3b)] is too The only ways of [a(a^2- 3b)] being a prime number is:

a=1 and [a^2 - 3b] is a prime number (1st hyphotesis) or [a^2 -3b] = 1 and a is a prime number.

1st hyphotesis : a = 1 and [1 - 3b] is a prime number. Impossible, because b is at least 1, so [1 - 3b] is no greater than (-2)

2nd hypohtesis: a is a prime number and [a^2 - 3b]=1 So 'a' has to be a prime, and a^2 = 3k +1 ; k is an integer a^2 - 3b = 3k +1 -3b = 0 => 3(k - b) = 0 => k = b a^2 = 3b +1 so a^2 is bigger than 4(3 +1) and smaller than 301(3.100 +1) squaring all the primes: 2^2 = 4 = 3.1 +1; It fits 3^2 = 9 = 3.3; It doesn't fit 5^2 = 25 = 3.8 +1; It fits 7^2 = 49 = 3.16 + 1; It fits 11^2 = 121 = 3.40 + 1; It fits 13^2= 169 = 3.56 +1; It fits 17^2 = 289 = 3.96 +1; It fits 19^2 = 361; It doesn't fit, because in this case a^2 is greater than 301.

Then, all the values of a are: {2,5,7,11,13,17}

So, the sum of possible values of a is: 2 + 5 + 7 + 11 + 13 +17 = 55

$x + y$ is a positive integer, therefore x and y are conjugates. if $x = k + ji$ ,

$a = 2k b = k^2 + j^2$

and

$x^3 + y^3 = 2k^3 - 6kj^3 = 2k(k^2 - 3j^2) = a(a^2 - 3b)$

For this to be a prime number, either one of the two factors must equal 1, and the other a prime. But clearly, $a$ cannot be 1; therefore $a$ must be prime and $a^2 - 3b = 1$ . Since all squares of primes (except 3) are $\equiv 1 \pmod{3}$ , this boils down to finding all primes whose squares are less than 301, i.e. 2, 5, 7, 11, 13, 17.

First, we observe that by the fundamental theorem of algebra, we have a pair of x and y for any a and b.

Now, $x^3 + y^3$ = $a^3 - 3ab$ . It is easy to determine that this is true by substituting in expressions for a and b.

So $a(a^2-3b)$ is prime. Hence, either $a = 1$ , so the expression is negative (a cannot be -1 since that is not in the required range, and 1-3b is negative for any valid b), or $a^2 - 3b = 1$ and a is prime.

Checking mod 3, we see that a is not 3, and noting that if $a^2 > 301$ , then $b > 100$ will hold, so the only values of a that will work are the primes less than 18 (other than 3). These are 2, 5, 7, 11, 13, 17 with a sum of 55.

We first note that $x^3+y^3=(x+y)(x^2-xy+y^2)=a((x^2+2xy+y^2)-3xy)=a(a^2-3b)$ . Since primes only have factors $1$ and $p$ , we know either $a=1$ and $1-3b=p$ (where p is a prime), or $a=p$ and $p^2-3b=1$ . Since no values of $b$ would give us $1-3b=p$ for a positive prime p, our only option is for $a=p$ and $p^2-3b=1$ . For $a=2, 5, 7, 11, 13, 17$ we have $b=1, 8, 16, 56, 96$ respectively. Larger prime values of $a$ will make $b>100$ .

Therefore, our answer is $2+5+7+11+13+17=55$ .

Factoring $x^3 + y^3$ , we have $(x+y)(x^2 - xy + y^2) = a(a^2-3b)$ . For this to be a positive prime, either $a$ must be 1 or $a^2 - 3b$ must be 1, and both must be positive integers.

$a = 1$ cannot be a solution because it makes $a^2 - 3b$ negative, so $a^2 - 3b$ must be 1, yielding ordered pairs $(2,1), (5,8), (7,16), (11,40), (13,56)$ , and $(17,96)$ for $a$ and $b$ . A quick check reveals that for any ordered pair $(a,b)$ , there exist complex number solutions $x$ and $y$ . Thus, our sum is $2 + 5 + 7 + 11 + 13 + 17 = 55$ .

x^3 + y^3 = (x+y)^3 - 3xy(x+y) = a^3 - 3ab = a(a^2 - 3b) which is divisible by a If x^3 + y^3 is a prime number then a=1 or a is a prime number and a^2-3b=1 Because prime numbers and b are positive so a can't be 1, then a must be prime and a^2 - 3b =1 (0< b < 101). Checking all the possible solutions we get a = 2,5,7,11,13,17 and 19

We notice that $x^{3}$ + $y^{3}$ = $(x + y)$ $(x^{2} + y^{2} - xy)$

= $(x + y)$ $((x + y)^{2} - 3xy)$

= $a$ $(a^{2} - 3b)$

Since $x^{3}$ + $y^{3}$ is a positive prime $\big($ Let's say P $\big)$ , so $a$ and $(a^{2} - 3b)$ are the factors of it. But we know the only factors a prime can have is $1$ and the prime itself. This leads to two cases:

$\bullet$ $a$ = $1$ and $(a^{2} - 3b)$ = P

$\implies$ P = $1 - 3b$ $<0$ $\big($ Since $3b$ $\geq3$ $\big)$ , which is unacceptable for a prime so rejected

$\bullet$ $(a^{2} - 3b)$ = $1$ and $a$ = P

Since $b$ $\leq100$

$\implies$ $a$ $\leq$ $\sqrt{301}$

Now ${17}^{2}$ $<$ $301$ $<$ ${18}^{2}$

$\implies$ $a$ $\leq17$ and because $a$ is a prime, so $a$ assumes all prime values between $2$ and $17$ $\big($ both inclusive $\big)$ . We should also keep in mind that $\forall$ prime values of $a$ found above, $b$ $\in$ $\{1, 2, 3, .........,100\}$ $\big($ This restriction eliminates $3$ as a possible value of $a$ $\big)$ .

So the sum we seek is $2 + 5 + 7 + 11 + 13 + 17$ = $\boxed{55}$