perplexus.info :: Just Math : Product + Square = Difference of Squares (2)

Product + Square = Difference of Squares (2) (Posted on 2011-07-20)

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.

Submitted by broll

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Solution: (Hide)

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)

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: Solution	broll	2011-07-26 02:08:50
Solution	Harry	2011-07-25 19:11:41

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