I'm looking for integer solutions to the equation $a^2+b^2+c^2=d^2$ such that $\gcd(a,b,c,d) = 1$.
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2These are known as pythagorean quadruples – lulu Nov 17 '18 at 19:49
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1If you just want a few ones, select $b$ and $c$ arbitrarily such that $b^2+c^2$ is odd, and let $a=\frac{b^2+c^2-1}2$, $d=a+1$. – hmakholm left over Monica Nov 17 '18 at 19:51
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Duplicate of https://math.stackexchange.com/q/2088960/451007. – Rócherz Nov 17 '18 at 19:53
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I've found the answer at https://math.stackexchange.com/questions/2088960/solutions-for-a2b2c2-d2 – Nov 17 '18 at 19:54
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1@Mason: The accepted answer does say that is describes the "primitive" solutions, which is another word for the gcd condition. If it is correct, then there are no pairwise coprime solutions. – hmakholm left over Monica Nov 17 '18 at 19:56
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1The problem as set has solutions with $c=0$ for example, which we already know. – Mark Bennet Nov 17 '18 at 19:58
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1Take any four integers $m,n,p,q$ with $\gcd(m,n,p,q)=1$ and $m+n+p+q$ odd. Those are the only rules. Then $$ a = m^2 + n^2 - p^2 - q^2, $$ $$ b = 2 (mq+np), $$ $$ c = 2(nq -mp), $$ $$ d = m^2 + n^2 + p^2 + q^2 $$ – Will Jagy Nov 17 '18 at 22:09