Primed Triangles

We have $16(p+15)=n(n+1)$ for some integer $n$ . Since $n$ and $n+1$ cannot both be even and $16 | n(n+1)$ , there is an integer $\tau$ such that either $n=16\tau$ or $n+1=-16\tau$ .

In either case, $16(p+15)=16\tau(16\tau+1)$ which can be rewritten as $p=\tau(16\tau+1)-15=(16\tau-15)(\tau+1)$ . One of those factors must equal $\pm1$ , since $p$ is prime, which leads to three (integral) possibilities: $\tau=1,p=2$ ; $\tau=-2,p=47$ ; or $\tau=0,p=-15$ which isn't prime.

Finally, verifying that $8(2+15)=16\cdot17/2$ and $8(47+15)=32\cdot31/2$ are in fact triangle numbers yields the conclusion that the sum of all prime numbers $p$ , such that $8p+120$ is a triangle number, is $2+47=49$ .

We have $8p+120 = \frac {n(n+1)}{2}$ , which gives $16p = n^2+n-240 = (n-15)(n+16)$ . Since $(n+16)-(n-15) =31$ is odd, $n+16$ and $n-15$ have different parity. Thus, 16 must divide one of the terms completely. Hence $\{ n-15, n+16 \} = \{ 16 p , 1 \}$ or $\{16, p \}$ . We consider the following cases:

Case 1: $n-15=1, 16p=n+16$ . Then $n=16$ , and $p=2$ .

Case 2: $n+16 =1, 16p=n-15$ . Then $n=-15$ , rejected as it is not a positive integer.

Case 3: $n-15=16, n+16=p$ . This yields $n=31, p=47$ .

Case 4: $n+16 = 16, n-15 = p$ . This yields $n=0$ , rejected as it is not a positive integer.

Hence, the only primes are $p=2, 47$ , which have sum $49$ .

A triangle number is of form n(n+1)/2, the sum of positive integers 1 to n. 120 is the sum of positive integers 1 to 15 = 15(15+1)/2, If 8p + 120 is sum ( 1+2+...+(n-1)+n), then 8p is sum ( 16+17+...+n). This is equal to (n-16+1)(16+n)/2. 16p = (n-15)(n+16). Clearly one of n-15 , n+16 is odd and the other even. The only odd factors of 16p are p and 1. (odd,even) pairs are (p,16) and (1,16p). It is trivial to note that (n-15, n+16) must be (16, p=47) and (1, 16p = 32). So p = 2 or 47, and their sum is 49.

 8p+120 = 8(p+15) = n(n+1)/2

or, 16(p+15) = n(n+1)

Hence, n(n+1) must contain 16 as a factor.

Hence, n or (n+1) must be a multiple of 16.

Let, n = 16k , or n+1 =16k k=1,2,3,………….

For k=1,

16(p+15) = 16 x 17 which is only possible because for 16 x 15, p=1(not prime)

Hence, p = 2.

For k=2,

16(p+15) = 31 x 32

Hence, p =47 which is only possible because for 32 x 33, p = 51 which is not a prime.

Further, it can be observed that no other combinations whose factor is 16 can be formed so that p comes out to be a prime. Hence, solutions are 2 and 47 Sum = 2+47=49

Since $8p+120= (n(n+1))/2.$ , Then after factoring, it becomes $(n+16)(n-15) = 16p$ . By further factoring, $16p$ is decomposed into five stuff, $1 * 16p$ $2 * 8p$ $4 * 4p$ $8 * 2p$ $16 * p$ . Then, difference in factor equals $(n+16) - (n-15) = 31$ By common senses, it is known that addition of odd number to an even number don't produce an even number. Since three cases in the middle have even coeficient for $n$ , $n-15$ require an even n so as to satisfy the first stuff $n+16$ as it should be divisble by the two. Thus shows a contradiction. So the three cases in the middle are abolished, So checking the first and last two cases show that $47, 2$ are the answer.

It turns out that there are only 2 primes that satisfy 8p+120 = a triangular number. Clearly, p = 2 works: 16+120 = 136 = 16 17/2. We will show that the only other value of p that works is p = 47. Thus the sum of all the required primes is 49. We need the following result: If n is a triangular number then 8n+1 is a square. So we can write the condition of the problem as 64p + 961 = x^2 or (x+31)(x-31) = 64p. Suppose p = 2 Since x > 31 we have x -31 = 2 x = 33, 128 + 961 = 1089 = 33^2. So let's assume p is an odd prime Then, because p is prime, we must have one of the following: a. x-31 = 32, x+31 = 2p Then x = 63, 63+31 = 94 94 32 + 961 = 3969, a square. So p = 47. 8p+120 = 496 = 31*32/2. b. x-31 = 16, x+31 = 4p x = 47. 47+31 = 78, but 78 is not of the form 4p. c). x-31 = 8. x+31 = 8p x = 39, 39+31 = 70, but 70 is not of the form 8p. Similarly, the remaining 2 cases are impossible. The answer to the problem is therefore 49.

It is clear that there are only 2 primes that satisfy 8p+120 = a triangular number. so, p = 2 works: 16+120 = 136 = 16 17/2. We will determine that the only other value of p that works is p = 47. so the sum of all the required primes is 49. We will have the following result: If n is a triangular number then 8n+1 is a square. So we can write the condition of the problem as 64p + 961 = x^2 or (x+31)(x-31) = 64p. Suppose p = 2 Since x > 31 we have x -31 = 2 x = 33, 128 + 961 = 1089 = 33^2. So let's suppose p is an odd prime Then, because p is prime, we must have one of the following: a. x-31 = 32, x+31 = 2p Then x = 63, 63+31 = 94 94 32 + 961 = 3969, a square. So p = 47. 8p+120 = 496 = 31*32/2. b. x-31 = 16, x+31 = 4p x = 47. 47+31 = 78, but 78 is not of the form 4p. c). x-31 = 8. x+31 = 8p x = 39, 39+31 = 70, but 70 is not of the form 8p. Similarly, the other 2 cases are impossible. The answer to the problem is therefore 49.

Only two numbers can solve this equation p = 2 works: 16+120 = 136 = 16 17/2. We need to show the only other value of p that works 47. So sum of all the primes that work is 49. We have to show the following : If n is a triangular number then 8n+1 is a square. So we can also say 64p + 961 = x^2 or (x+31)(x-31) = 64p. If p = 2 Since x > 31 we have x -31 = 2 x = 33, 128 + 961 = 1089 = 33^2. Let's predict p is an odd prime Then, because p is prime, we must have one of the following: a. x-31 = 32, x+31 = 2p Then x = 63, 63+31 = 94 94 32 + 961 = 3969, a square. So p = 47. 8p+120 = 496 = 31*32/2. b. x-31 = 16, x+31 = 4p x = 47. 47+31 = 78, but 78 is not of the form 4p. c). x-31 = 8. x+31 = 8p x = 39, 39+31 = 70, but 70 doesn't satisfy 8p. The remaining 2 cases don't works either. So 49 is the answer

It turns out that there are only 2 primes that satisfy 8p+120 = a triangular number. Clearly, p = 2 works: 16+120 = 136 = 16 17/2. We will show that the only other value of p that works is p = 47. Thus the sum of all the required primes is 49. We need the following result: If n is a triangular number then 8n+1 is a square. So we can write the condition of the problem as 64p + 961 = x^2 or (x+31)(x-31) = 64p. Suppose p = 2 Since x > 31 we have x -31 = 2 x = 33, 128 + 961 = 1089 = 33^2. So let's assume p is an odd prime Then, because p is prime, we must have one of the following: a.)x-31 = 32, x+31 = 2p Then x = 63, 63+31 = 94 94 32 + 961 = 3969, a square. So p = 47. 8p+120 = 496 = 31*32/2. b.) x-31 = 16, x+31 = 4p x = 47. 47+31 = 78, but 78 is not of the form 4p. c). x-31 = 8. x+31 = 8p x = 39, 39+31 = 70, but 70 is not of the form 8p. Similarly, the remaining 2 cases are impossible. The answer to the problem is therefore 49.

there are only 2 primes that satisfy 8p+120 = a triangular number. Clearly, p = 2 works: 16+120 = 136 = 16 17/2. so the first number is 2. If n is a triangular number then 8n+1 is a square. So we can write the condition of the problem as 64p + 961 = x^2 or (x+31)(x-31) = 64p. Suppose p = 2 Since x > 31 we have x -31 = 2 x = 33, 128 + 961 = 1089 = 33^2. So let's assume p is an odd prime Then, because p is prime, we must have one of the following: a. x-31 = 32, x+31 = 2p Then x = 63, 63+31 = 94 94 32 + 961 = 3969, a square. So p = 47. 8p+120 = 496 = 31*32/2. b. x-31 = 16, x+31 = 4p x = 47. 47+31 = 78, but 78 is not of the form 4p. c). x-31 = 8. x+31 = 8p x = 39, 39+31 = 70, but 70 is not of the form 8p. Similarly, the remaining 2 cases are impossible. The answer to the problem is therefore 49.

n(n+1)=16p+240

n^2 +n-240=16p

(n+16)(n-15)=16p

(n+16)(n-15) should be divisible by 16 and also the p should be prime n can be 16 thus p would be 2 or ncan be 31 p would be 47 there are no other possibilities so sum=47+2=49