Determine all possible positive integer triplet(s) (F, G, H) such that each of F and G is a prime with F ≤ G, and:

F(F+3) + G(G+3) = H(H+3)

F=3, G=7, H=8    and    F=2, G=3, H=4    are the only solutions.

Explanation:

(1)  F(F+3) + G(G+3) = H(H+3)

multiply by 4 and add 8FG  + 9 to each side (in the equations below I use LHS=left hand side, RHS=right hand side):

(2)  LHS = 4F^2 + 8FG + 4G^2 + 12F + 12G + 9 = (2F + 2G + 3)^2

(3)  RHS = 4H^2 + 12H + 9 + 8FG = (2H + 3)^2 + 8FG

subtract  (2H + 3)^2  from both sides and divide by 4:

(4)  (F + G - H) (F + G + H + 3)  = 2FG

Key point:  Since F and G are prime, there are two possibilties: Either the larger factor on the LHS of (4) equals 2FG, with the other factor equaling 1. Alternatively, one of the two factors on the LHS of (4) is identical with one of the factors 2, F or G, while the other factor equals the product of the other two terms.

There are thus seven cases to examine:

Case 1:  F + G - H = 1   and   F + G + H + 3 = 2FG

Case 2:  F + G - H = 2   and   F + G + H + 3 = FG

Case 3:  F + G - H = F   and   F + G + H + 3 = 2G

Case 4:  F + G - H = G   and   F + G + H + 3 = 2F

Case 5:  F + G - H = FG  and   F + G + H + 3 = 2

Case 6:  F + G - H = 2G  and   F + G + H + 3 = F

Case 7:  F + G - H = 2F  and   F + G + H + 3 = G

Cases 5-7 are obviously impossible as the second equation cannot hold.

In case 3 the first equation would require H=G, which is obviously incompatible with (1). Similarly, in case 4 the first equation would require H=F, another impossibility. This leaves cases 1 and 2 as the only remaining possibilities.

We examine case 1 by using the first equation to eliminate H, thus getting: 

(5.1)  F + G + 1 = FG

Note that there are no solution with F=G, as F^2 - 2F - 1 is positive for F>1. Therefore we assume G>F.

Next, suppose that F>=3. then FG >= 3G, while F + G + 1 <= 2G, an impossibility. It follows that we only need to consider G=2 with respect to case 1. Recapitulating (5.1) this leads to the sole case 1 type solution F=2,G=3,H=4

We examine case 2 in a similar fashion. By eliminating H in the first equation, we get

(5.2)  2F + 2G + 1 = FG

Here too, F=G does not work (as F^2 - 4F - 1 is negative for F<=4 and positive for F>4) so we again assume G>F.

Next suppose that F>=5, then FG > 4G while 2F + 2G + 1 < 4G, which is inconsistent with (5.2). Hence F would have to equal 2 or 3.

If F = 2 then (5.2) works out as 2G + 5 = 4G, which fails because the two sides have different parities. With F=3 we get G=7 and H=8 as the only case 2 type solution. 

Edited on April 19, 2008, 7:28 pm

Posted by FrankM on 2008-04-19 15:16:23