Đặt \(\left\{{}\begin{matrix}2\left(p+1\right)=k^2\\2\left(p^2+1\right)=m^2\end{matrix}\right.\)\(\left(k,m\in N,k,m>0\right)\)
Có: \(m^2-2p^2=k^2-2p\)\(\Leftrightarrow\left(m-k\right)\left(m+k\right)=2p\left(p-1\right)\)
*TH 1: \(\left\{{}\begin{matrix}m-k=2\\m+k=p\left(p+1\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m=k+2\\2k+2=p\left(p+1\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m=k+2\\2\left(k+1\right)=p\left(p+1\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m=k+2\\\left[{}\begin{matrix}\left\{{}\begin{matrix}p=2\\p=k\end{matrix}\right.\\\left\{{}\begin{matrix}p+1=2\\p=k+1\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow p=2;1\left(KTM\right)\)
Ttự xét TH: \(\left(m-k;m+k\right)=\left(2p;p-1\right);\left(p\left(p-1\right);2\right);\left(2\left(p-1\right);p\right)\);... để tính p.
Ta đặt :
\(\left\{{}\begin{matrix}\frac{p+1}{2}=a^2\\\frac{p^2+1}{2}=b^2\end{matrix}\right.\)(I)
\(\Rightarrow\left\{{}\begin{matrix}p+1=2a^2\\p^2+1=2b^2\end{matrix}\right.\)
\(\Rightarrow p\left(p-1\right)=2\left(b-a\right)\left(b+a\right)\)(*)
+) \(p=2\Rightarrow\) Vô lý
\(\Rightarrow p=2k+1\left(k\ge1\right)\)
\(\Rightarrow k\left(2k+1\right)=\left(b-a\right)\left(b+a\right)\)
\(\left(k;2k+1\right)=1\)
\(\Rightarrow\) Các trường hợp sau :
TH1 :
\(\left\{{}\begin{matrix}k\left(2k+1\right)=a+b\\b-a=1\end{matrix}\right.\)(II)
Từ (I) và (II) ta có :
\(\left\{{}\begin{matrix}b-a=1\\2a^2-1=\sqrt{2b^2-1}\end{matrix}\right.\)
Giải ta được : \(\left[{}\begin{matrix}a=0\\2a^3-3a-2=0\end{matrix}\right.\)
Ta thấy rằng :
\(a=0;a=1\)(loại)
\(\Rightarrow a\ge2\)
\(f\left(a\right)=2a^3-3a-2\)
\(\Rightarrow f'\left(a\right)=6a^2-3\)
\(\Rightarrow f\left(a\right)\) đồng biến trên [2;\(+\infty\))
\(\Rightarrow Min_{f\left(a\right)}=f\left(2\right)>0\)
\(\Rightarrow\) không có giá trị a thỏa mãn
TH2:
\(b-a\ge2\)
+)\(\left\{{}\begin{matrix}k=b-a\\2k+a=a+b\end{matrix}\right.\)
\(\Rightarrow3a-b=1\)(III)
Từ (I) và (III) ta có :
\(\left\{{}\begin{matrix}3a-b=1\\2a^2-1=\sqrt{2b^2-1}\end{matrix}\right.\)
Giải ta được :
\(\left[{}\begin{matrix}a=0\\a=2\end{matrix}\right.\)\(\Rightarrow a=2\)\(\Rightarrow p=7\)(t/m)
TH3:
Với \(\left(b+a\right)\left(b-a\right)|k\)
\(\Rightarrow k=i\left(b+a\right)\left(b-a\right)\left(i\ge1\right)\)
Thay vào (*) \(\Rightarrow i\left(2k+1\right)=1\Rightarrow2k+1\le1\Rightarrow k=0\)(Loại)
Với \(\left(b+a\right)\left(b-a\right)|\left(2k+1\right)\)
\(\Rightarrow2k+1=i'\left(b+a\right)\left(b-a\right)\left(i'\ge1\right)\)
Thay vào (*)\(\Rightarrow i'k=1\Rightarrow k\le1\)
\(\Rightarrow\)Loại
Vậy \(p=7\) thỏa mãn đề bài .
#Kaito#
