Problems & Puzzles: Puzzles

Puzzle 770. Eygptian Fraction Magic Square

Jon Perry sent the following nice puzzle:

Is it possible for a n*n square to be filled with 1/p's (p prime) such that each row/column (diagonal) has a subset such that the sum is 1?

Q. Please send your best solution, if any.

Contributions came from Giovanni Resta and Ramón David Aznar.


Giovanni wrote:

 if the primes
can be repeated the puzzle is trivial since
1/2+1/2=1 and thus a 2x2 square full of 1/2

If the primes cannot be repeated, then there is not
a solution since an Egyptian fraction summing up to 1
cannot contain only primes.
This is very easy to see. I can show this for 4 primes
and the proof by contradiction
generalize trivially to any number of terms.

Suppose you have 4 primes p,q,r,s and that
1/p+1/q+1/r+1/s = 1, then putting together the
fractions we have
(pqr+pqs+prs+qrs)/(pqrs)=1 or
pqr+pqs+prs+qrs = pqrs. Now I collect p:

p(qr+qs+rs) + qrs = pqrs and I divide by p:

qr + qs + rs + qrs/p  = qrs, but this is impossible
because all the terms are integers, apart from
qrs/p that cannot be integer because q,r,s, and p by
hypothesis are primes. So the sum cannot be 1.


Ramón wrote:

Si los valores de P son diferentes no es posible.

Supongamos que existe una solución 1/P1+1/P2+...+1/Pn=1. Tenemos entonces que la suma de los primeros n-1 términos es una fracción cuyo denominador (no reducido) es el producto de los primeros n-1 valores primos de P. Como esa suma es 1-1/Pn=(Pn-1)/Pn, el denominador de dicha fracción, Pn, puede ser escrito como el producto de los primeros n-1 valores primos de P. Pero como Pn es coprimo con todos ellos, no es posible.

Si los valores de P pueden ser iguales la solución trivial está compuesta por fracciones 1/N en un cuadrado NxN.


Records   |  Conjectures  |  Problems  |  Puzzles