A triplet (1,25,49) represents 3 perfect squares forming an arithmetic progression.

1. Provide few additional samples like the above.
2. What can be said about the possible values of d (the constant
difference between the adjacent members of an arithmetic progression)?
3. How relates the title of my post to its contents?

This answer deals only with 'regular' solutions.

Let the 3 numbers to be squared be a, b, c. By basic algebra:

a^2+d=b^2 
a^2+2d=c^2 
From 1 
a^2+2(b^2-a^2)=c^2 
2b^2=c^2+a^2 

This will always be true when a=b=c, but assume a and b are different.

a can be any integer we choose, and there will always be a solution:
5a=b, 7a=c, d=(5^2-1)a^2  
29a=b, 41a=c, d=(29^2-1)a^2  
169a=b, 239a=c, d=(169^2-1)a^2, etc.

Generally, let x be a number such that 2x^2-1=y^2 has a solution in the positive integers. Then a, b=(ax), c=(ay), d=(x^2-1)a^2 is a solution:

(ax)^2-a^2 = (ay)^2-(ax)^2
a^2(x^2-1) = a^2(y^2-x^2)

Clearing the a^2 terms and collecting the others: 2x^2-1=y^2;  the relationship between x and y is independent of a and is fixed.

As I said above, 'regular' solutions, because a might, e.g. be 7, in which case {b,c,d} = {13,17,120}{17,23,240} etc. also represent valid, but not 'regular' solutions.

I'm not sure about the 'UN cargo' clue, though. If it is a clue, that is...

 

 

 

Edited on March 13, 2013, 12:21 pm

Posted by broll on 2013-03-13 12:11:52