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This topic contains 2 replies, has 3 voices, and was last updated by Tarit Goswami 1 month, 2 weeks ago.

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May 28, 2018 at 10:10 pm #21566
The problem says: A 7tuple \((a_1,a_2,a_3,a_4,b_1,b_2,b_3)\) of pairwise distinct positive integers with no common factor is called a shy tuple if \(a_1^2+a_2^2+a_3^2+a_4^2=b_1^2+b_2^2+b_3^2\) and for all \(1\le i<j\le 4\) and \(1\le k\le 3\) , \(a_i^2+a_j^2\ne b_k^2\) . Prove that there exists infinitely many shy tuples.
My solution: Take \((a_1,a_2,a_3,a_4,b_1,b_2,b_3)=(m^2+n^2,12k,9k,36k, m^2n^2, 2mn, 39k)\) , where \(m\) is odd, \(n\) is even and \(k\) is any integer that makes each of the 7 numbers distinct. Since we have odd and even numbers in the tuple, they have no common factor. It is easy to verify that \((m^2+n^2)^2+(12k)^2+(9k)^2+(36k)^2=(m^2n^2)^2+( 2mn)^2+( 39k)^2 \). There are infinitely many such \(m,n,k\), so infinitely many such 7tuples.
Could someone please check whether this solution works.
PS: I assumed that it is not mandatory that the numbers are pairwise coprime as the problem does not say so.
 This topic was modified 3 months, 3 weeks ago by Writika Sarkar. Reason: latex
July 10, 2018 at 11:20 pm #21683first of all i am assuming that numbers are not supposed to be pairwise co prime so your solution need more work on m,n,k . You have to show what exact condition we need on k to get perfect solution out of it.Here I have different approach.
now take a2=3 a3=4 a4=12 and b3=13 and now we know that we have infinite pythagorean triples so we will take (a1,b1,b2) as a pythagorean triple. where a1^2=b1^2 + b2^2 wlog assume that b1>b2 so we take all pythagorean triples with b2>13 as condition.
July 30, 2018 at 11:39 pm #21802Yes, Shantanu is correct, they want to say,
pairwise distinct and no common factor
i.e; it is mendetory to consider that they are paurwise coprime. Otherwise, it is pointless to say that the all seven have “no common factor”, when they have said that they are “pairwise distinct”. I will come back with solution. Give it a try again.
 This reply was modified 1 month, 2 weeks ago by Tarit Goswami.
 This reply was modified 1 month, 2 weeks ago by Tarit Goswami.

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