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\(Page.48\\1.\ 如果形如\ 4k-1\ 的素数有无穷多个,设为4k_{1}-1,4k_{2}-1,···,4k_{n}-1\\\quad 设其余小于4k_{n}-1但不包含2的所有素数为4k‘_{1}+1,4k‘_{2}+1,···,4k‘_{m}+1\\\quad 则c=2*(4k_{1}-1)···(4k_{n}-1)(4k‘_{1}+1)···(4k‘_{m}+1)+1为素数且可以表达为4k‘+1的形式\\\quad 并且有2\not\mid (4k_{1}-1)···(4k_{n}-1)(4k‘_{1}-1)···(4k‘_{m}-1)即2\not\mid \prod_{i=1}^{n}(4k_{n}-1)\ *\ \prod_{j=1}^{m}(4k_{m}+1)\\\quad 2*\prod_{i=1}^{n}(4k_{n}-1)\ *\ \prod_{j=1}^{m}(4k_{m}+1)+1=4k‘+1\\\quad \prod_{i=1}^{n}(4k_{n}-1)\ *\ \prod_{j=1}^{m}(4k_{m}+1)=2k\\\quad 则可得到\ 2\mid \prod_{i=1}^{n}(4k_{n}-1)\ *\ \prod_{j=1}^{m}(4k_{m}+1)\ ,与假设矛盾,故形如\ 4k-1\ 的素数有无穷多个\)?
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\(2.\ 用数学归纳法易证,下面给出更有趣(并不)的论断的证明\\\quad 设p_{i}为素数,p_{i}=4k\pm 1,n=p_{1}p_{2}···p_{n},则n可表示为4k\pm 1的形式,n^{2}可表示为4k+1的形式\\\quad 或n=2p_{1}p_{2}···p_{n},则n可表示为4k的形式,n^{2}可表示为4k的形式\\\quad 设q_{i}为素数,q_{i}=4k\pm 1,m=q_{1}q_{2}···q_{m},则m可表示为4k\pm 1的形式,m^{2}可表示为4k+1的形式\\\quad 故任意m^{2}+n^{2}\ (g.c.d(m,n)=1)\ 可表达为4k+1的形式\\~\\\quad 当m^{2}+n^{2}为一个素数时,\ p=m^{2}+n^{2}=4k+1\ 显然成立\\~\\\quad 当m^{2}+n^{2}为一个合数时,设r\mid m^{2}+n^{2}\ ,sr=m^{2}+n^{2}\\\quad 则必然存在0\leqslant m_{1}<r,0\leqslant n_{1}<r,\ s.t.\ m_{1}\equiv m(mod\ \ r),\ n_{1}\equiv n(mod\ \ r)\\\quad 注意若m_{1}=0,则r\mid m,此时r\mid n^{2}=sr-m^{2},又r为素数,故r\mid n,与\ g.c.d(n,m)=1\ 矛盾\\\quad 故存在0<m_{1}<r,0<n_{1}<r,\ s.t.\ m_{1}\equiv m(mod\ \ r),\ n_{1}\equiv n(mod\ \ r)\\\quad 即m_{1}^{2}\equiv m^{2}(mod\ \ r),\ n_{1}^{2}\equiv n^{2}(mod\ \ r),\ m_{1}^{2}+n_{1}^{2}\equiv m^{2}+n^{2}\equiv 0(mod\ \ r)\\\quad 故可设s_{1}r=m_{1}^{2}+n_{1}^{2},g.c.d(m_{1},n_{1})=t_{1}\\\quad 即t_{1}\mid m_{1},\ t_{1}\mid n_{1},\ t_{1}^{2}\mid m_{1}^{2}+n_{1}^{2},\ t_{1}^{2}\mid s_{1}r\\\quad 设左右同除t_{1}^{2}后得s_{2}r=m_{2}^{2}+n_{2}^{2},g.c.d(m_{2},n_{2})=1\\\quad 当\ s_{2}=1\ 时,r=m_{2}^{2}+n_{2}^{2}可以表达为4k+1的形式,讨论当s_{2}>1时的情况\\~\\\quad 若s_{2}=2k,则有2\mid m_{2}^{2}+n_{2}^{2},即m_{2},n_{2}同奇偶\\\quad 设s_{3}=\frac{s_{2}}{2},\ m_{3}=\frac{\left | m_{2}+n_{2}\right |}{2},\ n_{3}=\frac{\left | m_{2}-n_{2}\right |}{2}\\\quad 由\frac{s_{2}}{2}r=(\frac{\left | m_{2}+n_{2}\right |}{2})^{2}+(\frac{\left | m_{2}-n_{2}\right |}{2})^{2}得,\ s_{3}r=m_{3}^{2}+n_{3}^{2},\ g.c.d(m_{3},n_{3})=t_{3},由g.c.d(m_{2},n_{2})=1可知g.c.d(m_{3},n_{3})=t_{3}=1\\~\\\quad 若s_{2}=2k+1>1,对于s_{2}r=m_{2}^{2}+n_{2}^{2}\\\quad 则必然存在0\leqslant m_{3}<s_{2},0\leqslant n_{3}<s_{2},\ s.t.\ m_{3}\equiv m_{2}(mod\ \ s_{2}),\ n_{3}\equiv n_{2}(mod\ \ s_{2})\\\quad 注意若m_{3}=0,则s_{2}\mid m_{2},此时s_{2}\mid n^{2}=s_{2}r-m^{2},与\ g.c.d(n_{2}^{2},m_{2}^{2})=1\ 矛盾\\\quad 故存在0<m_{3}<s_{2},0<n_{3}<s_{2},\ s.t.\ m_{3}\equiv m_{2}(mod\ \ s_{2}),\ n_{3}\equiv n_{2}(mod\ \ s_{2})\\\quad 即m_{3}^{2}\equiv m_{2}^{2}(mod\ \ s_{2}),\ n_{3}^{2}\equiv n_{2}^{2}(mod\ \ s_{2}),\ m_{3}^{2}+n_{3}^{2}\equiv m_{2}^{2}+n_{2}^{2}\equiv 0(mod\ \ s_{2})\\\quad 故可设\ s_{3}s_{2}=m_{3}^{2}+n_{3}^{2}\ ,又s_{2}r=m_{2}^{2}+n_{2}^{2}\\\quad 有s_{2}^{2}s_{3}r=(m_{3}^{2}+n_{3}^{2})(m_{2}^{2}+n_{2}^{2})=(m_{2}m_{3}+n_{2}n_{3})^{2}+(m_{2}m_{3}-n_{2}n_{3})^{2}\\\quad 又\ m_{2}m_{3}\equiv m_{3}^{2}(mod\ \ s_{2})\quad \ n_{2}n_{3}\equiv n_{3}^{2}(mod\ \ s_{2})\quad \ m_{2}m_{3}-n_{2}n_{3}\equiv 0(mod\ \ s_{2})\\\quad 即s_{2}^{2}\mid (m_{2}m_{3}-n_{2}n_{3})^{2}\quad s_{2}^{2}\mid (m_{2}m_{3}+n_{2}n_{3})^{2}\\\quad 得s_{3}r=(\frac{m_{2}m_{3}+n_{2}n_{3}}{s2})^{2}+(\frac{m_{2}m_{3}-n_{2}n_{3}}{s2})^{2}\\\quad 故可设\ m_{4}=\frac{\left | m_{2}m_{3}+n_{2}n_{3}\right |}{s2}\ n_{4}=\frac{\left | m_{2}m_{3}-n_{2}n_{3}\right |}{s2}此时s_{3}r=m_{4}^{2}+n_{4}^{2}\quad g.c.d(m_{4},n_{4})=t_{4}\\\quad 可得t_{4}\mid m_{4},t_{4}\mid n_{4},t_{4}^{2}\mid m_{4}^{2},t_{4}^{2}\mid n_{4}^{2}\\\quad 对\ s_{3}r=m_{4}^{2}+n_{4}^{2}\ 左右同除\ t_{4}^{2}\ 得\ s_{4}r=m_{5}^{2}+n_{5}^{2} \quad g.c.d(m_{5},n_{5})=1\ 若此时s_{4}=1,r=m_{5}^{2}+n_{5}^{2}可以表达为4k+1的形式\\\quad 若此时s_{4}>1,则根据能依靠上述方法再次讨论,直至s_{4}=1时结束.\\~\\\quad Q.E.D\)
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\(3.\ 数学归纳法易证\)
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\(4.\ 数学归纳法易证\)
原文：https://www.cnblogs.com/cihua/p/Dsxyl_1_1_9.html