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Problem 785


Problem 785


Symmetric Diophantine equation

Consider the following Diophantine equation:
15(x2+y2+z2)=34(xy+yz+zx)
where x, y and z are positive integers.

Let S(N) be the sum of all solutions, (x,y,z), of this equation such that, 1xyzN and gcd(x,y,z)=1.

For N=102, there are three such solutions - (1,7,16),(8,9,39),(11,21,72). So S(102)=184.

Find S(109).


对称丢番图方程

考虑如下丢番图方程:
15(x2+y2+z2)=34(xy+yz+zx)
其中xyz均为正整数。

对于上述方程满足1xyzNgcd(x,y,z)=1的解(x,y,z),记S(N)为所有这些解之和。

对于N=102,共有三组这样的解:(1,7,16),(8,9,39),(11,21,72),因此S(102)=184

S(109)


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