May 28, 2018 at 10:10 pm #21566
The problem says: A 7-tuple \((a_1,a_2,a_3,a_4,b_1,b_2,b_3)\) 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 \((a_1,a_2,a_3,a_4,b_1,b_2,b_3)=(m^2+n^2,12k,9k,36k, m^2-n^2, 2mn, 39k)\) , 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 \((m^2+n^2)^2+(12k)^2+(9k)^2+(36k)^2=(m^2-n^2)^2+( 2mn)^2+( 39k)^2 \). There are infinitely many such \(m,n,k\), so infinitely many such 7-tuples.
Could someone please check whether this solution works.
PS: I assumed that it is not mandatory that the numbers are pairwise co-prime as the problem does not say so.
July 10, 2018 at 11:20 pm #21683
- This topic was modified 3 months, 3 weeks ago by Writika Sarkar. Reason: latex
first 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 #21802
Yes, Shantanu is correct, they want to say,
pairwise distinct and no common factor
i.e; it is mendetory to consider that they are paurwise co-prime. 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.
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