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Problem 730

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Shifted Pythagorean Triples

For a non-negative integer $k$, the triple $(p,q,r)$ of positive integers is called a $k$-shifted Pythagorean triple if $$p^2+q^2+k=r^2$$

$(p,q,r)$ is said to be primitive if $gcd(p,q,r)=1$.

Let $P_k(n)$ be the number of primitive k-shifted Pythagorean triples such that $1\le p\le q\le r$ and $p+q+r\le n$.
For example, $P_0({10}^4)=703$ and $P_{20}({10}^4)=1979$.

Define
$${\displaystyle S(m,n)=\sum _{k=0}^m{P}_k(n)}$$
You are given that $S(10,{10}^4)=10956$.

Find $S({10}^2,{10}^8)$.

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移位勾股数

对于非负整数$k$，若正整数三元组$(p,q,r)$满足
$$p^2+q^2+k=r^2$$
则称之为$k$-移位勾股数。若进一步地$(p,q,r)$满足$gcd(p,q,r)=1$，则称为$k$-移位本原勾股数。

记$P_k(n)$为满足$1\le p\le q\le r$和$p+q+r\le n$的$k$-移位本原勾股数的数目。
例如，$P_0({10}^4)=703$，$P_{20}({10}^4)=1979$。

记
$${\displaystyle S(m,n)=\sum _{k=0}^m{P}_k(n)}$$
已知$S(10,{10}^4)=10956$。

求$S({10}^2,{10}^8)$。

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