Prime Numbers

Number Theory Level 5

Find all positive integers $n > 1$ such that $\frac{ n(n+1) } { 2} - 1$ is prime.

Enter your answer as the sum of the number of such positive integers and all of the corresponding prime values.

The answer is 9.

1 solution

William Isoroku
Nov 15, 2016

Let $n=2a$ , then the equation becomes:

$\frac { n(n+1) }{ 2 } -1\Longrightarrow \frac { 2a(2a+1) }{ 2 } -1\Longrightarrow \frac { 2a(2a+1)-2 }{ 2 }$ which then simplifies to $a(2a+1)-1$

We want this to equal to a prime $p$ ;

$a(2a+1)-1=p\Longrightarrow 2a^2+a-(1+p)=0$

Use the quadratic formula:

$a=\frac { -1+\sqrt { 1+8(p+1) } }{ 4 }$ , (we neglect the negative values because we want positive $n$ , thus must have positive $a$ )

Solve for the discriminant: $1+8(p+1)=c^2\Longrightarrow8(p+1)+1=c^2$ for some positive integer $c$ because we want the discriminant to be a perfect square all the while $a$ also has to be an integer or a fraction that yields an integer when multiplied by 2

Simple trial and error gives us $p=2,5$ and there are 2 values for $n$

Note: we can also approach the last part by solving for $c^2\equiv 1 (\mod{8})$ by using quadratic residues.

Better yet, we can just use factorization. We have,

$\frac{n(n+1)}2-1=p\iff n(n+1)=2(p+1)\iff n^2+n-2=2p\iff (n+2)(n-1)=2p$

Now, since $n\gt 1$ , we have $n-1\in\{1,2,p,2p\}$ corresponding to $n+2\in\{2p,p,2,1\}$ from which we can note that only the first two cases give sensible values for $(n,p)$ and they are $(2,2)$ and $(3,5)$ .

Prasun Biswas - 4 years, 7 months ago

You should add this as a solution.

Chaebum Sheen - 4 years, 7 months ago

Looks like I just like to complicate things

William Isoroku - 4 years, 7 months ago

Prime Numbers

The answer is 9.

1 solution

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