Replace the numerals 1 through 8 with ever increasing prime numbers, always using the next lowest possible that is available1
to fulfill the criteria on the left. Then do the same for the right.
Present a series for the left, and one for the right.
|7 + 8 = Square
6 = Prime
4 + 5 = Square
1 + 2 + 3 = Square
|7 + 8 = Cube
6 = Prime
4 + 5 = Cube
1 + 2 + 3 = Cube
If it was required that the Right set required "1" to be the next Prime following on after that used for the "8" in the Left set, what might the Right set read, if indeed it is possible?
Note, "always using the next lowest possible that is available"
means that if it is next on the list it cannot be dismissed unless it is the last
of a group of two or three and will not fulfill the criterion. Only then may you advance to the next.
(In reply to continuing on with the "square" version (spoilers for part 1)
I'm sure there are other much more elegant ways to prove this, plus many other existing sequences that can be used to explain why this is, but I was bored and decided to think for myself for a change.
The values you've listed as resulting in no solutions are simply pairs of consecutive primes that add up to a perfect square.
The two starting primes (P1, P2) must sum to an even value, so let's set x=y² where x is the sum, and y is the root of the perfect square.
Now, adding some prime (P3) to the mix and getting a higher square (this time it is always an odd number), leads to the following.
x+P3=(y+a)² where a is any odd positive integer. This will expand to:
Take out the x=y², and we're left with P3=2ay+a². This factors to a(2y+a), so any value for a other than 1 will result in a composite number.
So, P3=1(2y+1), or P3=2y+1. As x is the sum of P1 and P2, let x=2z where z is the average of P1 and P2.
From this, y=√(2z), which leads to an integral value for z=2, and z=8. As z can never be 2 (no two primes average to 2), our minimum value for z must be 8.
When z=8, x=16, and y=4. At this point, x=2z=4y, or z=2y.
That means that for all primes (P1, P2) that add to a perfect square, P3=2y+1≤P2, and thus no valid value for P3 will exist.
Edited on January 7, 2010, 2:29 am
Posted by Justin
on 2010-01-07 02:14:48