Hiển nhiên \(c\left(c+1\right)>a\left(a+1\right)\Rightarrow c>a\ge b\)
Nếu \(c\ge2a\Rightarrow c\left(c+1\right)\ge2a\left(2a+1\right)=4a^2+2a\)
Mà \(a\left(a+1\right)+b\left(b-1\right)\le a\left(a+1\right)+a\left(a-1\right)=2a^2\)
\(\Rightarrow2a^2\ge4a^2+2a\Rightarrow2a^2+2a\le0\) (vô lý)
\(\Rightarrow c< 2a\)
Ta có:
\(4a\left(a+1\right)+4b\left(b-1\right)+1=4c\left(c+1\right)+1\)
\(\Leftrightarrow4a\left(a+1\right)+\left(2b-1\right)^2=\left(2c+1\right)^2\)
\(\Leftrightarrow4a\left(a+1\right)=\left(2c+1\right)^2-\left(2b-1\right)^2\)
\(\Leftrightarrow a\left(a+1\right)=\left(c-b+1\right)\left(c+b\right)\) (*)
Nếu \(c-b+1\ge a\Rightarrow\left(c-b+1\right)\left(c+b\right)>a\left(a+b\right)>a\left(a+1\right)\) (ktm)
\(\Rightarrow c-b+1< a\) \(\Rightarrow c-b+1\) ko có ước nguyên tố nào là a
\(\Rightarrow c+b⋮a\Rightarrow\dfrac{c+b}{a}\in Z\) (1)
Theo chứng minh ban đầu, ta có \(b\le a< c< 2a\)
\(\Rightarrow a< c+b< 2a+a=3a\Rightarrow1< \dfrac{c+b}{a}< 3\) (2)
(1);(2) \(\Rightarrow\dfrac{c+b}{a}=2\Rightarrow c+b=2a\)
Thế vào (*) \(\Rightarrow a+1=2\left(c-b+1\right)\Rightarrow2c-2b+1=a\)
\(\Rightarrow2\left(2a-b\right)-2b+1=a\Rightarrow3a-4b+1=0\)
\(\Rightarrow3\left(a-1\right)=4\left(b-1\right)\)
\(\Rightarrow b-1⋮3\Rightarrow b-1=3k\Rightarrow b=3k+1\)
\(\Rightarrow a=4k+1\)
\(\Rightarrow c=2a-b=5k+1\)
\(\Rightarrow A=3\left(5k+1\right)-5\left(3k+1\right)=-2\)
