F(F+3)+G(G+3)=H(H+3) where F and G are both primes and
F≤G
=> 3(F+G-H)=(H²-F²-G²)
Let F and G be primes other than 3
then F² mod 3 = G² mod 3 = 1
H² mod 3 = 0 or 1
So, (H²-F²-G²) mod 3 = 2 or 1
In this case, RHS is not divisible whereas LHS is
divisible by 3. So, atleast one of F,G must be 3.
case(1): Let F=3
=> 3*6=H²+3H-G²-3G
=> 18 = (H-G)*(H+G+3)
We know that H>G≥F{=3} => H+G+3>9
The only divisor of 18 greater than 9 is 18
So, H+G+3=18 and H-G=1 => G=7 and H=8
{3,7,8} is one triplet.
case(2): Let G=3
F can be 2 or 3 according to given conditions
If F=3
H²+3H-36=0 No integer solutions to this quadratic
since its discriminant 153 is not a perfect square
and If F=2
H²+3H-28=0 => (H+7)(H-4)=0
As H is positive, H=4
{2,3,4} is another triplet that satisfies the given
equation.
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Posted by Praneeth
on 2008-04-22 07:34:53 |
Solution
