Pretty Primes

Number Theory Level 5

Find all solutions in primes $p \leq q \leq n$ to the equation

$p(p+1) + q(q+1) = n(n+1)$

Enter your solution as $\sum p_i + q_i + n_i$ .

The answer is 7.

1 solution

Aditya Narayan Sharma
Feb 22, 2016

Our equation yields, p(p+1) = n(n+1) - q(q+1) p(p+1) = (n-q)(n+q+1) .............................. (1) Now, We must have n>q.

For p is a prime ,

Case 1 : p | (n-q) Case 2: p | (n+q+1)

Proceeding with case 1 we have,

$p \le (n-q)$ which implies that, $p(p+1) \le (n-q)(n-q+1)$ But from (1) we get,

$(n+q+1) \le (n-q+1)$ which is inadmissible , so we cancel this case.

For Case 2 :

p | (n+q+1) So, n+q+1 = ap (For some a) Therefore,

p(p+1) = (n-q)(n+q+1)

p(p+1) = (n-q)ap

p+1 = (n-q)a .................................................. (2)

If we had a = 1, (n+q+1) = p & (p+1) = n-q, which gives p-q = n+1 & p+q = n-1, which is impossible. p+q > p-q n - 1 > n + 1 which is impossible . Thus, a > 1.

From (2) we easily obtain,

2q = (n+q)-(n-q)

= ap-1-(n-q)

= a[a(n-q)-1]-I-(n-q)

= (a+1)[(a-l)(n-q)-I] ......................... (3)

Now since $a \ge 2$ , $(a+1) \ge 3$

From (3) we see that the divisors of L.H.S are 1,2,q,2q only.

So if (a+1) = q .

(a-1)(n-q) - 1 = 2

(a-1)(n-q) = 3

(q-2)(n-q) = 3 , which leads to either ,

q-2 = 1 or n-q = 3

So,
Here q=3,n=6,a=2

And from n+q+1=ap,

we see that p=5.

Again , if (q-2)=3 & (n-q)=1

We get q=5,n=6,k=4 & p=3

On the other hand if (a+1) = 2q,

(a-1)(n-a) = 2 2(q-1)(n-a) = 2 (q-1)(n-a) = 1

So, q-1 = 1 & n-a = 1

We get

q=2,n=3 and p=2 from (n+q+1) = ap.

Thus we observe that there is only one set of solution in primes for the above equation.

p=q=2 & n=3

p+q+n=2+2+3 = 7

Moderator note:

I do not understand how

So, n+q+1 = ap (For some a)
Therefore,
p+1 = (n-q)a (2)

(In the next line, you want "a=1" instead of "k=1").

From there on, I'm unable to follow your solution.

Good answer

Sayandeep Ghosh - 5 years, 3 months ago

I do not understand how

So, n+q+1 = ap (For some a)
Therefore,
p+1 = (n-q)a (2)

(In the next line, you want "a=1" instead of "k=1").

From there on, I'm unable to follow your solution.

Calvin Lin Staff - 5 years, 3 months ago

Sorry sir, I have edited that 'k' as 'a' and that line :

we have p | (n+q+1) so n+q+1 = ap (For some integer a)

putting it in,

p(p+1) = (n-q)(n+q+1) p(p+1) = (n-q)ap So we get , p+1 = (n-q)a .

Aditya Narayan Sharma - 5 years, 3 months ago

Ah ic. Please add that step in. Otherwise, from $n+q + 1 = ap$ , I only conclude that $p = \frac{n+q+1}{a}$ . Now I understand the rest of your solution.

I think that it can be greatly simplified, by using the symmetric condition that $q \mid n+p+1$ . We have $n < p + q$ and $n+p+1 \leq 4q$ , from which considering the values of $a_q = 1, 2, 3, 4$ could be easier.