smci
Lv 7
smci asked in Science & MathematicsMathematics · 1 decade ago

Find n integers s.t. the sum or abs difference of any two of them is a square?

Find n distinct integers {a,b,c...} such that the sum or absolute difference of ANY two of them is a square.

Obviously this will be possible for n=2., by picking

a = (m²+k²)/2 , b = (m²-k²)/2

(Find examples)

Also, prove whether it's possible for n=3,4 or arbitrarily large n.

(Examples for n=3, 4?)

This is a generalization of Dragan's question:

http://answers.yahoo.com/question/index?qid=200910...

Update:

[JB] Yes I meant 'both sum and absdiff are squares'. Also a,b,c... must be positive integers.

7,6,3,2 doesn't work as 7+6 is not a square, neither are 7+3, 6+2 or 3+2.

For n distinct integers, there will be n(n-1)/2 distinct sums and n(n-1)/2 distinct absdiffs, and they must ALL be squares, it's a pretty tight constraint, and it's quadratic in n.

I was looking for people to e.g. generalize the approach

a = (m²+k²)/2 , b = (m²-k²)/2

to a set of 3 or more integers.

Update 2:

[JB] I tried my best to be clear...

2 Answers

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  • JB
    Lv 7
    1 decade ago
    Best answer

    7, 6, 3, 2 works for four integers, since

    7-6 = 1^2, 7-3 = 2^2, 7+2 = 3^2, 6+3 = 2^2, 6-2 = 2^2, 3-2 = 1^2

    Or did you mean 'and' in your question, instead of 'or' ?

    Even simpler 3, 2, 1, -1, or do you want only positive integers?

    ---- ---- -----

    With the problem as originally stated I also found:

    4002, 2082, 2066, 1298, 638, 622

    where

    4002+2082 = 78^2, 4002-2066 = 44^2, 4002-1298 = 52, 4002-638 = 58^2, 4002+622 = 68^, 2082-2066 = 4^2, 2082-1298 = 28^2, 2082-638 = 38^2, 2082+622 = 52^2, 2066+1298 = 58^2, 2066+638 = 52^2, 2066-622 = 38^2, 1298+638 = 44^2, 1298-622 = 26^2, 638-622 = 4^2.

    Also 5437, 1837, 1788, 1212, 813, 188 where I won't write down sums or differences which are squares, since that was not the intended problem.

    --- ---- ---- ----

    With the new problem as explained in additional details, well, I will have to start fresh ...

    LATER: OK, triples exist, but are not easy to find. Here are three of them:

    [a,b,c] = [434657, 420968, 150568]. Then

    [a+b, a-b, a+c, a-c, b+c, b-c] = [925^2, 117^2, 765^2, 533^2, 756^2, 520^2]

    [a,b,c] = [2843458, 2040642, 1761858]. Then

    [a+b, a-b, a+c, a-c, b+c, b-c] = [2210^1, 896^2, 2146^2, 1040^2, 1950^2, 528^2]

    [a,b,c] = [3713858, 891458, 88642]. Then

    [a+b, a-b, a+c, a-c, b+c, b-c] = [2146^2, 1680^2, 1950^2, 1904^2, 990^2, 896^2]

    These were found by using parameters that sum to squares in two ways, causing the first 4 quantities to be squares automatically, so only square b+c and b-c are searched for.

    I do not know if quadruples exist. If so, they will be hard to find.

    --- ---- ---- ----

    ADDITIONAL DETAILS:

    put a = (L^2+M^2)/2; b = (L^2-M^2)/2, c = -(L^2+M^2)/2 + U^2, where L,M,U are parameters. Then:

    [a+b, a-b, a+c, a-c] = [L^2, M^2, U^2, L^2+M^2-U^2], so if L^2+M^2 = U^2+V^2 then

    [a+b, a-b, a+c, a-c] = [L^2, M^2, U^2,V^2], so if the search space is numbers n such that n is a sum of squares in two ways, say n = L^2+M^2 = U^2+V^2, then we only need to search for both b+c and b-c to be squares.

  • 1 decade ago

    the latest calculator could help: casio fx millenium

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