Problems & Puzzles:
Puzzles
Puzzle 123. Two
Frank Rubin's prime
puzzles
On my request Frank Rubin kindly sent
(29/12/2000) - among others - the following two puzzles for these pages:
I.- "What
is the largest number of (A) distinct integers, or (B) distinct
primes, such that the sum of any 3 of them is prime?"
II.- "Show
that there can not be a triangle in which the 3 sides and the area are
all primes"
_______
N.B. : I just want to say tonight
that Frank invents delicious puzzles.
He certainly is a Puzzler for Puzzles-Makers.
Solution
I.- Solutions to this Puzzle came form Jud McCranie,
Jean-Charles Meyrignac, J. Brette & Phil Carmody
The larger set of integers is a set with 5 (five)
members. The smallest of these sets is {-9, 3, 9, 11, 17}.
The largest set of primes is a set with 4 (four)
members. The smallest of these sets is {5, 7, 17, 19}
All proofs sent goes something like this:
"For the case of all primes, then, you can't
have more than 4 numbers, and their residues mod 3
must be (1,1,2,2). One solution is (7,11,13,23).
For the case of arbitrary integers the subset which
is a multiple of 3 could be 3 itself, so 5 numbers are possible. Their
residues mod 3 can be either (0,0,1,1,1) or (0,0,2,2,2). One solution is
(-5,1,7,15,21)" (F.R)
II.- Solutions to this puzzle came from Luis
Rodríguez, Felice Russo & Phil Carmody
The Rodríguez proof goes something like this:
"A^2 = area = S(S-a)(S-b)(S-c), where S = (a+b+c)/2
A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16 = Z/16
Let c<=b<=a. There are 4 (four) cases if a, b & c are
primes, that cover all the situations:
1) c = b = a = 2: Then A^2 = 6, then A is not prime
2) c = b = 2, a = 3: Then A^2= 7x1x3x3/16, then A is not prime
3) c = 2: if only c=2 then or b=a or b<a. But a<=b+2 in a triangle.
Then b=a =p > 2, then A^2 = (p-1).(p+1) =p^2-1 then A is
not a square number and consequently A is not a prime.
4) c>2: Then (b&a) >2, then Z = odd; then A is not prime.
Conclusion: if (a, b & c) are primes then A is not prime.
Conclusion: It's not possible that (a, b, c & A) are all primes"
(C.R.)
The Russo's proof has a different
approach:
"Let's indicate with p1, p2 and p3 (where
p1,p2 and p3 are any prime) the 3 sides of the triangle and with h its
height. Moreover let's suppose that p3=p3a+p3b where p3a and p3b are the
sides of the two 90degree triangle (p3a,h,p1) and (p3b,h,p2).
Let's suppose now that the area of the triangle is a prime number
p. Then A=(p3*h)/2=p that imply:
1) h=2 and p3=p or
2) h=p and p3=2
because A must be of course integer and prime.
Let's consider now the case 1). Applying the Pythagoras theorem we
have:
p2^2-2^2=p3b^2 --> p2^2-p3b^2=4 that is absurd because the
difference between two squared integers is never 4. So A cannot be a
prime
Now let's consider the case 2). By using again the Pythagoras
theorem we have:
p2^2-h^2=p3b^2
But p3b^2<p3^2=4 and then p2^2-p^2<4 but the difference of
two squared primes is greater than or equal to 5 . So A cannot be a
prime.
This prove completely the problem. But we can generalize this
result in the following one using the same argumentations used above:
The area of any triangle with the 3 sides integers cannot be a prime."
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