Let Q(x)=Q(x1,x2,⋯,xn)\documentclass[12pt]{minimal}
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\begin{document}$$Q(\mathbf{{x}}) = Q(x_1 ,x_2 ,\dots ,x_n )$$\end{document} be a nonsingular quadratic form over Z\documentclass[12pt]{minimal}
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\begin{document}$$\mathbb {Z}$$\end{document}, and p\documentclass[12pt]{minimal}
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\begin{document}$$p$$\end{document} be an odd prime. A solution of the congruence Q(x)≡0(modpm)\documentclass[12pt]{minimal}
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\begin{document}$$Q({\mathbf {x}}) \equiv {\mathbf {0}}\,(\mathrm{mod}\, p^m )$$\end{document} is said to be a primitive solution if p∤xi\documentclass[12pt]{minimal}
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\begin{document}$$p\not \mid x_i $$\end{document} for some i\documentclass[12pt]{minimal}
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\begin{document}$$i$$\end{document}. We prove that if p>A,\documentclass[12pt]{minimal}
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\begin{document}$$p > A,$$\end{document} where A=22(n+1)/(n-2)32/(n-2)\documentclass[12pt]{minimal}
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\begin{document}$$ A = 2^{2(n + 1)/(n - 2)} 3^{2/(n - 2)}$$\end{document}, then this congruence has a primitive solution, with x≤61/np(m/2)+(m/n)\documentclass[12pt]{minimal}
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\begin{document}$$ \left\| \mathbf{{x}} \right\| \le 6^{1/n} p^{(m/2) + (m/n)}$$\end{document} whenever n>m\documentclass[12pt]{minimal}
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\begin{document}$$n>m$$\end{document} and m≥2,\documentclass[12pt]{minimal}
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\begin{document}$$m\ge 2,$$\end{document} for every even n\documentclass[12pt]{minimal}
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\begin{document}$$n$$\end{document}.
