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Greatest common divisor of the product of terms in positive Pythagorean n-tuples, over all possible such tuples.
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%I #23 Jun 21 2024 15:20:05

%S 60,12,2,24,6,8,12,3,4,12,6,8,6,24,4,3,12,4,6,24,2,24,12,1

%N Greatest common divisor of the product of terms in positive Pythagorean n-tuples, over all possible such tuples.

%C A Pythagorean n-tuple consists of positive integers X1,...,Xn where X1^2 + X2^2 + ... + X(n-1)^2 = Xn^2. The product of those terms is P = X1*X2*...*Xn.

%C a(n) is the GCD of all possible products P arising this way.

%C If n is odd, then 2 | a(n).

%C If n mod 3 is 0 or 1, then 3 | a(n).

%C If n is divisible by 4, then 4 | a(n).

%C If n mod 8 is 1 or 3, then 4 | a(n).

%C If n mod 8 is 0 or 6, then 8 | a(n).

%C If n mod 24 is 2, then a(n)=1.

%C a(n+24)=a(n) for n >= 4.

%H Des MacHale and Christian van den Bosch, <a href="https://www.jstor.org/stable/23249521">Generalising a result about Pythagorean triples</a>, The Mathematical Gazette, Vol. 96, March 2012.

%e For n = 8, then 8 | a(n). Since 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 = 4^2 with P = 1*1*1*1*2*2*2*4 = 32 and 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2 = 5^2 with P = 1*1*1*1*1*2*4*5 = 40, then a(8) = gcd(32,40) = 8, and no larger number will divide the product of terms in every Pythagorean octuple.

%K nonn,more

%O 3,1

%A _Brian Almond_, May 28 2024