Find the sum of the triple primes which fulfill these conditions

Number Theory Level 3

{ p q = r + 1 2 ( p 2 + q 2 ) = r 2 + 1 \begin{cases} p q = r + 1 \\ 2(p^2 + q^2) = r^2 + 1 \end{cases}

Let ( p , q , r ) (p,q,r) be a triplet of prime numbers satisfying the system of equations above.

If all the unordered triplets of the solutions of ( p , q , r ) (p,q,r) are ( p 1 , q 1 , r 1 ) , … , ( p n , q n , r n ) (p_1, q_1, r_1) , \ldots , (p_n, q_n , r_n) , submit your answer as ∑ m = 1 n ( p m + q m + r m ) \displaystyle \sum_{m=1}^n (p_m + q_m + r_m) .


The answer is 10.

Vijay Simha by Vijay Simha , 47, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Vijay Simha
Nov 9, 2017

If p and q are both odd, then r = pq - 1 is even so r= 2.

But in this case pq >= 3*3 = 9 and hence there are no solutions.

This proves that either p = 2 or q = 2.

If p = 2 then we have 2*q = r + 1

and

8 + 2q^2 = r^2+ 1.

Multiplying the second equation by 2 we get 2r^2+ 2 = 16 + (2q)^2 = 16 + (r+ 1)^2.

Rearranging the terms, we have r^2-2r-15 = 0,or equivalently (r+ 3)(r-5) = 0.

This proves that r = 5 and hence q = 3.

Similarly,if q= 2 then r= 5 and p= 3.

Thus the only two solutions are (p;q;r) = (2;3;5) and (p;q;r) = (3;2;5)

Since we are supposed to discard the triple if any of the numbers are repeated, in the second triple, r is repeated, so it can be discarded.

So our sum is 2 + 3 + 5 = 10.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...