$\frac{b-1}{a-1}=\frac{b+1}{a+1}+1$ $(b-1)(a+1)=(a-1)(a+b+2)$ $ab-a+b-1=a^2+ab+2a-a-b-2$ $a^2+2a+1=2b+2$ $(a+1)^2=2(b+1)$ ,so $2| a+1$ Similar we get $(b+1)^2=2(c+1)$ or $(a+1)^4=8(c+1)$ $8*150+24 \geq 8(a+1)+8(b+1)+8(c+1)=8(a+1)+4(a+1)^2+(a+1)^4$ $(a+1)^4+4(a+1)^2+8(a+1) \leq 1224 \to a+1 < 6 $ Case 1: $a+1=2 \to b+1=2$ - contradiction Case 2: $a+1=4 \to b+1=8 \to c+1=32$ $(a,b,c)=(3,7,31)$

How did u assume that b>a

Because$\frac{b-1}{a-1}$ is integer

Oops sorry