perplexus.info :: Just Math : Square Sum in Pairs

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

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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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