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 Square Sum in Pairs (Posted on 2016-06-27)
(i) X and Y are two positive integers such that:
Each of 2016+X, 2016+Y and X+Y is a perfect square.

Find the two smallest values of X+Y

(ii) If in addition to given conditions in (i), it is known that X is itself a perfect square and Y is two times a perfect square- then, what is the smallest value of X+Y?

 No Solution Yet Submitted by K Sengupta No Rating

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 not too much Comment 2 of 2 |
I don't know a good way for solving this puzzle out of trying.

Anyway what I've done is assign a parameter (r, p, q) to each of 2016, X, Y, so that

2016+X=(r+p)^2  [1]
2016+Y=(r+q)^2  [2]
X+Y=(p+q)^2  [3]

From [3]-[2]-[1] can obtain:

2016=r(r+p+q)-pq

If, for example, we suppose that X, Y are lower than 2016 (and then p, q, lower than r), we minimize the effect of -pq. Parameter r will be in the interval (30,45) and p, q in the interval (1,30)

This can be explore with Excel.

r       p     q       X        Y         X+Y

32   16    30     288    1828    46^2
24    28   1120    1584    52^2
33     6    27     -495    1584    33^2
15    24     288    1233    39^2
35     4    21     -495    1120    25^2
36     4    18     -416      900    22^2
12    12     288      288    24^2
39     1    12     -416      585    13^2
40     3      8     -167      288    11^2

The negative values verify that 2016-X, 2016+Y, X+Y are squares.

Edited on June 30, 2016, 9:28 am
 Posted by armando on 2016-06-30 09:26:33

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