An algebra problem by Krishna Sharma

Consider $c=apq<a$ .

Then $b=a (p+q) <apq+a=c+a$ as $a (1-p)(1-q) > 0$ .

Also $b > 2\sqrt {pq} a=2\sqrt {ac}$ . This is by AM-GM inequality.

So, now $b$ is bounded between two numbers. One of them, $c+a$ is an integer. This means that the other, $2\sqrt {ac} <c+a-1$ or equivalently $\sqrt {a}-\sqrt {c}> 1$ . This is by bringing $2\sqrt {ac}$ across and square rooting.

(Otherwise $b$ would be between 2 consecutive integers, impossible)

As $c \geq 1$ , this immediately implies $a>4$

Thus the minimum is at least 5. But the quadratic $5x^{2}-5x+1$ satisfies the condition. Thus 5 is the answer.

D=b²-4ac>0 because p,q are real roots,so b²>4ac. Becuse of 0<(b±√(b²-4ac))/2a<1 ,we get -b<±√(b²-4ac)<2a-b ,first inequality is alwaz true because -b<0 and -4ac<0 so we get condition √(b²-4ac)<2a-b which means 2a-b>0 and b²-4ac<4a²-4ab+b², so b<2a and -4ac<4a²-4ab, when divided by 4a means -c<a-b <=> b<a+c. So we have 3 conditions: b<2a,b+<a+c,b²>4ac. To get minimum value for a we must guessing from 1 . a=1 our conditions become
b<2,b<a+c,b²>4ac .Obious b must be 1 from first inequality,but then b²=1 and 1>4c is not true because c is positive integer. So a is not 1.For a=2 we get b<4,b<2+c,b²>8c max for b is 3 so b² max is 9 and for c=1 b²=9>8=4ac,but b<a+c=3 is not satisfied so a is not 2.For a=3 we get b<6,b<3+c,b²>12c. For c=1 b must be at least 4 but then b<a+c=4 is not satisfied.Same for c=2.b must be 5,but then b<a+c=5 is not satisfied.For c=3 b must be at least 6,but how b<6 that can not be the case,so a is not 3.For a=4 we get b<8,b<4+c,b²>16c,for c=1 b<a+c=5 but then b²>16 is not satisfied,so lets try c=2,then b<4+c=6 but then b²>16 is not satisfied again,so for c=3 b<4+c+7 ,but b²>48 is not satisfied,so for c=4 b<4+c=8 but b²>64 is not satisfied,for c=5 because of condition b<2a=8 inequality b²>80 is not satisfied and for c>5 right side of inequality is getting bigger so a=4 is not solution. FOr a=5, our conditions become b<2a=10 , b<5+c , b²>20c and for c=1 they became b<10,b<6,b²>20 ,so b=5 is solution. MINIMUM POSSIBLE VALUE OF a IS 5 and only one equation satisfy our conditions 5x²-5+1, roots are in that case (5±√(25-20))/10 , roots are 1/2±1/(2√5), p~0,7236,q~0,2764