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Product + Square = Difference of Squares (2) (Posted on 2011-07-20) Difficulty: 3 of 5
Following on from this problem, three positive integers P, Q and R, (from smallest to largest in that order), are in arithmetic sequence satisfying : N*P*Q*R + Q = R^2 - P^2, where N is a positive integer.

Determine all possible quadruplet(s) (P, Q, R, N) that satisfy the above equation, and prove that no other quadruplet satisfies the given conditions.

Note that in this variant, the second term involving Q is not a square.

See The Solution Submitted by broll    
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re: Solution Comment 2 of 2 |
(In reply to Solution by Harry)

Much neater than my solution! However,

Write as n*(p-k)*p*(p+k) + p = (p+k)^2 - (p-k)^2  
np(p-k)(p+k)+p = 4kp  
n(p-k)(p+k)+1 = 4k  
n(p^2-k^2)+1 = 4k  
k=1,n=1,p=2 is the only solution.

Proof of uniqueness is more troublesome. However if n is of the form (4m+1) then {k,p}={2s-1,2r} when s quickly exceeds r, and if n is of the form (4m+3) then then {k,p}={2s,2r-1} when all (fractional) solutions are given by:

s -(m+3)/(4m+3), (m+1)/(4m+3)  
hence k=2s={-2(m+3)/(4m+3), (2m+2)/(4m+3)   
r (m)/(4m+3), 3(m+1)/(4m+3)  
hence p=2r-1={(-2m+3)/(4m+3), (2m+3)/(4m+3)

 

 


  Posted by broll on 2011-07-26 02:08:50
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