RMO 2016 Delhi Problem 5 a)

[Writika Sarkar]

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.

[shantanu deodhar]

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.

 

[Tarit Goswami]

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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