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In this paper we reveal the connection between quartic residues and the sum $\\sum_{k=0}^{[p/4]}\\binom{4k}{2k}\\frac 1{m^k}$, where $[x]$ is the greatest integer not exceeding $x$. Let $q$ be a prime of the form $4k+1$ and so $q=a^2+b^2$ with $a,b\\in\\Bbb Z$. 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