Given p is a prime number and that there are 2 distinct positive integers u and v such that p^2 is the mean of u^2 and v^2.

Prove that 2p−u−v is either a square or twice a square.

It's not really to do with primes. The starting point is the insight that the values of p are always of the form (x^2+y^2), which happens to yield some small primes at the outset.

Once this is grasped, it is readily seen that (as one possible formulation) v= 2xy+(y^2-x^2),   p=(x^2+y^2),  u=2xy+(x^2-y^2) or u=(y^2-x^2)-2xy, since: 

(2xy+(y^2-x^2))^2+(2xy+(x^2-y^2))^2 =
(2xy+(y^2-x^2))^2+((y^2-x^2)-2xy)^2 =  2(x^2 + y^2)^2

But then:

2(x^2+y^2)-(2xy+(y^2-x^2))-(2xy+(x^2-y^2)) = 2(x-y)^2.

2(x^2+y^2)-(2xy+(y^2-x^2))-((y^2-x^2)-2xy) = (2x)^2.

So 2p-u-v is either a square or twice a square, as was to be shown.

NOTE: Since p^2 is a mean, we might want to know the common difference between the pairs of squares. It is a cubic: 4x^3y - 4xy^3, or 4xy(x-y)(x+y), always divisible by 24, a result first noted by Fibonacci in Liber Quadratorum.

Edited on January 15, 2017, 1:23 am

Posted by broll on 2017-01-15 00:27:32