求所有的正整數數對$，滿足$a|b^2,b|a^2,(a+1$(a,b))|(b^2+1)$

求所有的正整數數對\((a,b)\)，滿足\(a|b^2,b|a^2,(a+1)|(b^2+1)\)

解
設\(d=(a,b),a=da_1,b=db_1\)，則有\((a_1,b_1)=1\)
則 \(a|b^2\Rightarrow da_1|d^2b_1^2\Rightarrow a_1|db_1^2\Rightarrow a_1|d\)
同理可知 \(b_1|d\)，故\(a_1b_1|d\)，並且設\(d=d_1a_1b_1\)
代入第三式得

\[\begin{align} &d1a_1^2b_1+1|d_1^2a_1^2b_1^4+1\\ \Leftrightarrow &d_1a_1^2b_1+1|d_1^2a_1^2b_1^4+1-d_1b_1^3(d_1a_1^2b_1+1)\\ \Leftrightarrow &d_1a_1^2b_1+1|d_1b_1^3-1 \end{align} \]

\((i)\)

當 \(d_1b_1^3-1=0\)時，\(d_1=b_1=1\)，有解 \(b=a_1,a=a_1^2\)
\((ii)\)當 \(d_1b_1^3-1\ne 0\)時，\(d_1b_1^3-1\ge d_1a_1^2b_1+1\Leftrightarrow d_1b_1^3>d_1a_1^2b_1\Rightarrow b_1>a_1\)
\(d_1a_1^2b_1+1|(d1b_1^3-1)a_1^2-(d1a_1^2b_1+1)b_1^2=-a_1^2-b_1^2\)
所以 \(d_1a_1^2b_1+1|a_1^2+b_1^2\)
故 \(a_1^2+b_1^2\ge a_1^2b_1+1\Rightarrow b_1^2-1\ge a_1^2(b_1-1)\Rightarrow b_1+1\ge a_1^2\)
又 \(d_1a_1^2b_1+1|(a_1^2+b_1^2)d_1a_1^2-b_1(d_1a_1^2b_1+1)=d_1a_1^4-b_1\)
故有第二組解 \(b_1=d_1a_1^4\)
若 \(d_1a_1^4-b_1<0\)，則 \(|d_1a_1^4-b_1|<b_1<d_1a_1^2b_1+1\)
若 \(d_1a_1^4-b_1>0\)，則 \(d_1a_1^4-b_1\ge d_1a_1^2b_1+1\)
由 \(d_1a_1^2(b_1+1)=d_1a_1^2b_1+d_1a_1^2\) 與 \(b_1+1\ge a_1^2\)
知 \(d_1a_1^4-b_1<d_1a_1^4< 2(d_1a_1^2b_1+1)\)
故由整除可知

\[\begin{align} d_1a_1^4-b_1&=d_1a_1^2b_1+1\\ d_1a_1^4b_1-1&=(d_1a_1^2+1)(a_1^2-1)\\ &=d_1a_1^4+a_1^2-da_1^2-1 \end{align} \]

故第三組解有 \(d_1=1,b_1=a_1^2-1\)
綜上所述，當 \(t,d\in \mathbb{Z}^+\) 時, \((t^2,t),(d^3t^6,d^3t^9),(t^4-t^2,t^5-2t^3+t)\) 都是合法的解

本文來自TYNFMS，作者：畢天馳，轉載請註明原文連結：https://www.cnblogs.com/tynfms-bjq/p/15369409.html