a) The quadratic equation $x^2 + (a + b - 1)x + ab - a - b = 0$ has two distinct real solutions if its discriminant is positive, that is, $(a+b-1)^2 - 4(ab-a-b) > 0$. Simplifying this expression, we get $(a-b+1)^2 > 0$, which is true for all $a\neq b-1$. Since $a\leq b$, we have $a\leq b-1$ only if $a=b-1$, which means $a$ and $b$ are consecutive positive integers. However, this is not possible because the quadratic equation would then reduce to $x^2=0$, which has only one root. Therefore, for all $a$ and $b$ with $a\leq b$, the equation has two distinct real solutions. b) Let $r$ and $s$ be the two solutions of the quadratic equation $x^2 + (a + b - 1)x + ab - a - b = 0$. If one of the solutions is an integer, say $r$, then we have $r+s=a+b-1$ and $rs=ab-a-b$. From the first equation, we get $s=a+b-1-r$. Substituting this into the second equation and simplifying, we get $r^2-(a+b-1)r+ab-a-b=0$. This is again a quadratic equation in $r$ with solutions $r$ and $s$, so we must have $s=\frac{ab-a-b}{r}$. Since $r$ is an integer, it follows that $s$ is also an integer. Furthermore, since $r$ is a solution of the quadratic equation, we have $r^2 + (a + b - 1)r + ab - a - b = 0$. Substituting $s=a+b-1-r$ into this equation and simplifying, we get $(r-a)(r-b+1)=0$. Since $a\leq b$, we have $r=a$ or $r=b-1$. If $r=a$, then $s=b-1$, and if $r=b-1$, then $s=a$. In either case, we have $r+s=a+b-1$, which implies that both $r$ and $s$ are non-positive integers. To show that $b<2a$, suppose for contradiction that $b\geq 2a$. Then $a+b-1 \geq 3a-1$, and we have $r+s=a+b-1\geq 3a-1$. Since $r$ and $s$ are non-positive integers, it follows that $r\leq -a$ and $s\leq -2a+1$, so $rs\geq 2a^2-a$. But this contradicts the fact that $rs=ab-a-b<a(b-1)-a-b=a(b-2)-b<2a^2-a$, since $b-2\geq a$. Therefore, we must have $b<2a$.

a) $(a+b-1)^2-4(ab-a-b)=(b-a+1)^2+4a>0$ so equation has two distinct real solutions b) If $m,n$ are roots and $m$ is integer then $n=-a-b+1-m$ is integer too If $a=1$ then equation $x^2+bx-1=0$ can not have integer solutions because $D=b^2+4$ is not a square of integer. So $a \geq 2,b \geq 2$ $m*n = ab-a-b=(a-1)(b-1)-1 \geq 0$ and $m+n=-(a+b-1) \leq -3$ so both $m,n$ have same sign and so they both are non-positive. $D=(b-a+1)^2+4a$ is square of positive integer and so $(b-a+1)^2+4a \geq (b-a+3)^2 \to b \leq 2a-2$