Integrally rational

We know that the roots of $ax^2 + bx + c$ are $\frac {-b+sqrt{b^2-4ac}} {2a}$ and $\frac {-b- sqrt{b^2-4ac}}{2a}$ . Now it can easily be proved that one of these numbers is rational if and only if $b^2-4ac= k^2$ for some integer $k$ . The given problem can now be divided into 5 cases:- $b=1, b=2, b=3, b=4$ , and $b=5$ .

Case 1 $b=1$
The equation now becomes $1-4ac= k^2$ . This obviously has no solution over the integers since $1-4ac\leq 1-4*1*1=-3<0$ , but the square of a real number has to be non-negative.

Case 2 $b=2$
The equation now becomes $4-4ac= k^2$ , or $4(1-ac)= k^2$ . Since 4 is a perfect square and so is $k^2$ , $1-ac$ must also be a perfect square. Since $1-ac\leq 1$ (a and c being positive integers) and $k^2\geq 0$ , the only possible value for $1-ac$ is 0, which implies $ac=1$ . This has one solution over the integers, which is $(a, c)= 1$ . So case 2 gives one solution over the integers.

Case 3 $b=3$
The equation now becomes $9-4ac=k^2$ . Since $9\equiv1\pmod{4}$ and $4ac\equiv0\pmod{4}$ so $k^2\equiv1\pmod{4}$ . Also $9-4ac\leq 9-4*1*1= 5$ . The only perfect square less than 5 and congruent to 1mod 4 is 1. So $9-4ac= 1$ , or $ac= 2$ . This has two solutions over the integers, which are (a, c)= (1, 2) and (a, c)= (2,1). So case 3 gives two solutions over the integers.

Case 4 $b=4$
The equation now becomes $16-4ac= k^2$ , or $4(4-ac)= k^2$ . Since 4 is a perfect square and so is $k^2$ , $4-ac$ must also be a perfect square. This gives 2 possible values for ac:- $ac=4$ and $ac=3$ . The case $ac=4$ has 3 solutions over the integers, and the case $ac=3$ has 2 solutions over the integers. So case 4 gives 5 solutions over the integers.

Case 5 $b=5$
The equation now becomes $25-4ac= k^2$ . Since $25\equiv1\pmod{4}$ and $4ac\equiv1\pmod{4}$ , $k^2\equiv1\pmod{4}$ . Also $k^2<25$ . The only perfect squares less than 25 and congruent to 1 mod 4 are 9 and 1. The case $k^2=9$ gives $ac= 4$ , which has 3 solutions over the integers. The case $k^2=1$ gives ac=6, which has 2 solutions over the integers, since we are restricted to $1 \leq a, c, \leq 5$ . So case 5 gives 5 solutions over the integers.

Note that no solutions will overlap from different cases since the value of $b$ is different in different cases. So adding we get that the number of solutions is 13.

[LaTeX edits - Calvin]

Note that the quadratic formula for finding solutions of x is $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . So in order for the solutions of x to be rational, the portion $\sqrt{b^2-4ac}$ has to be rational, since $a, b$ and $c$ are integers. Thus $\sqrt{b^2-4ac}$ has to be an integer.

[The following cases are long, but well written and explained - Calvin]

Case 1: b=1. Then $b^2-4ac=1-4ac$ . Since $a,c \ge 1$ , thus $1-4ac \le 1-4=-3$ . But $b^2-4ac$ has to be non-negative, else $\sqrt{b^2-4ac}$ is imaginary. So NO SOLUTIONS for this case.

Case 2: b=2. Note that $b^2-4ac=4-4ac \le 4-4=0$ . But $b^2-4ac$ has to be non-negative. So the only rational solutions are when $b^2-4ac=4-4ac=0$ . Solving gets $ac=1$ . Since $a,c \ge 1$ , thus the only possible solution is when a=1 and c=1. Thus the only possible triplet (a,b,c) that is possible is $(1,2,1)$ .

Case 3: b=3. If ac=1, then $b^2-4ac=5$ and $\sqrt{b^2-4ac}=\sqrt{5}$ . But that would mean that the solutions for x are irrational. So no solutions for ac=1. If ac=2, then $b^2-4ac=1$ and $\sqrt{b^2-4ac}=1$ . Note that the solutions for ac=2 are a=1,c=2 or a=2,c=1. So the possible triplets for this sub-case are $(1,3,2)$ and $(2,3,1)$ . If ac=3 or more, then $b^2-4ac\le-3$ and $\sqrt{b^2-4ac}$ would be imaginary. So no solutions for ac=3 or more.

Case 4: b=4 If ac=1, then $b^2-4ac=12$ and $\sqrt{b^2-4ac}=\sqrt{12}$ . But that would mean that the solutions for x are irrational. So no solutions for ac=1. If ac=2, then $b^2-4ac=8$ and $\sqrt{b^2-4ac}=\sqrt{8}$ . But that would mean that the solutions for x are irrational. So no solutions for ac=2. If ac=3, then $b^2-4ac=4$ and $\sqrt{b^2-4ac}=2$ . Note that the solutions for ac=3 are a=1,c=3 or a=3,c=1. So the possible triplets for this sub-case are $(1,4,3)$ and $(3,4,1)$ . If ac=4, then $b^2-4ac=0$ and $\sqrt{b^2-4ac}=0$ . Note that the solutions for ac=4 are a=1,c=4 or a=4,c=1 or a=2,c=2. So the possible triplets for this sub-case are $(1,4,4)$ and $(4,4,1)$ and $(2,4,2)$ . If ac=5 or more, then $b^2-4ac\le-4$ and $\sqrt{b^2-4ac}$ would be imaginary. So no solutions for ac=5 or more.

Case 5: b=5 If ac=1, then $b^2-4ac=21$ and $\sqrt{b^2-4ac}=\sqrt{21}$ . But that would mean that the solutions for x are irrational. So no solutions for ac=1. If ac=2, then $b^2-4ac=17$ and $\sqrt{b^2-4ac}=\sqrt{17}$ . But that would mean that the solutions for x are irrational. So no solutions for ac=2. If ac=3, then $b^2-4ac=13$ and $\sqrt{b^2-4ac}=\sqrt{13}$ . But that would mean that the solutions for x are irrational. So no solutions for ac=3. If ac=4, then $b^2-4ac=9$ and $\sqrt{b^2-4ac}=3$ . Note that the solutions for ac=4 are a=1,c=4 or a=4,c=1 or a=2,c=2. So the possible triplets for this sub-case are $(1,5,4)$ and $(4,5,1)$ and $(2,5,2)$ . If ac=5, then $b^2-4ac=5$ and $\sqrt{b^2-4ac}=\sqrt{5}$ . But that would mean that the solutions for x are irrational. So no solutions for ac=5. If ac=6, then $b^2-4ac=1$ and $\sqrt{b^2-4ac}=1$ . Note that the solutions for ac=6 are a=2,c=3 or a=3,c=2 (Note that a=1,c=6 or a=6,c=1 are not solutions since a,c<=5). So the possible triplets for this sub-case are $(2,6,3)$ and $(3,6,2)$ . If ac=7 or more, then $b^2-4ac\le-3$ and $\sqrt{b^2-4ac}$ would be imaginary. So no solutions for ac=7 or more.

In conclusion, the $13$ solutions are: (1,2,1), (1,3,2), (2,3,1), (1,4,3), (3,4,1), (1,4,4), (4,4,1), (2,4,2), (1,5,4), (4,5,1), (2,5,2), (2,5,3), (3,5,2).

The solutions of a second degree equation are x1=(−b+√D)/2a and x2=(−b−√D)/2a. In order to have a rational solution D has to be a positive square. D= b^2-4ac>=0. Now i rosolve this equation for b bein 1, 2, 3, 4 and 5. For b=1 i have no positive quarter. For b=2 i have a positive quarter (0) for a and c being 1 and 1 therefore (1,2,1). For b=3 I have a positive quarter for a and c being (2,1) and (1,2): (1,2,2) and (2,2,1). For b=4 i have positive quarters for a and c being (2,2) (3,1) (1,3) (4,1) (1,4): (2,4,2) (3,4,1) (1,4,3) (4,4,1) (1,4,4). For b=5 i have positive quarter for for a and c being (2,2) (3,2) (2,3) (4,1) (1,4): (2,4,2) (3,4,2) (2,4,3) (4,4,1) (1,4,4). Counting the different triples i optain 13.

Notice that $a \neq 0$ , we have a quadratic with rational solutions. Thus the discrminant, $b^2-4ac$ , must be a perfect square. Let $b^2-4ac=k^2$ , where $k$ is a nonnegative integer.

Then

$(b-k)(b+k)=4ac$

Notice that $b-k$ and $b+k$ has the same parity, so they must be both even. Then

$(\frac{b-k}{2})(\frac{b+k}{2})=ac$

where $\frac{b-k}{2}$ and $\frac{b+k}{2}$ are both integers. Let $\frac{b-k}{2}=x, \frac{b+k}{2}=y$ . Since $y$ is a positive integer and $ac$ is positive, $x$ must also be a positive integer. We have

$xy=ac, x+y=b$

where $a$ and $c$ are positive integers smaller or equal than $5$ . Also

$5 \geq b=x+y \geq 2 \sqrt{xy}=2 \sqrt{ac} \Rightarrow ac \leq (\frac{5}{2})^2 \Rightarrow ac \leq 6$

since $ac$ is an integer.

If $ac=1$ , we must have $x=y=a=c=1$ , and $b=2$ . 1 triple.

If $ac=2$ , we could have $a=1, c=2$ or $a=2, c=1$ . And $b=1+2=3$ . 2 triples.

If $ac=3$ , we could have $a=1, c=3$ or $a=3, c=1$ . And $b=1+3=4$ . 2 triples.

If $ac=4$ , we could have $a=1, c=4; a=2, c=2$ or $a=4, c=1$ . And $b=1+4=5$ or $2+2=4$ . Thus we have 6 triples.

If $ac=5$ , we could have $a=1, c=5$ or $a=5, c=1$ . However $b$ could only be $1+5=6$ , which is too large.

If $ac=6$ , we could have $a=2, c=3$ or $a=3, c=2$ . And $b=2+3=5$ . 2 triples.

Summing up, we have $1+2+2+6+2=13$ triples.

The roots of the quadratic equation are $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ , and the part under the square root is the only cause of irrational numbers.

So, $b^2-4ac$ is a perfect square $\iff$ $ax^2+bx+c=0$ has a rational solution.

Hence, we try from b=1 to b=5 to yield the following solutions:

• b=2; a,c=1,1 (1 solution)
• b=3; a,c=1,2 (2 solutions)
• b=4; a,c=1,3 (2 solutions)
• b=4; a,c=2,2 (1 solution)
• b=4; a,c=1,4 (2 solutions)
• b=5; a,c=2,2 (1 solution)
• b=5; a,c=1,4 (2 solutions)
• b=5; a,c=2,3 (2 solutions)
• b=5; a,c=1,6 (This solution is rejected as $a,c\leq5$ )

Note that this usage of a,c is irrespective of order, hence 2 solutions when $a\neq c$ .

Hence the total number of solutions is $13$ .

If the quadratic has a rational root, then the other root must also be rational since all the coefficients are positive integers.

We will determine when the discriminant of the quadratic is a perfect square, which will mean that the quadratic will have rational roots.

Case 1: $a = 1, 2≤b≤5$

Discriminant is $b^2 - 4c$ .

We know that 2≤b≤5, since the discriminant must be positive.

The discriminant is a perfect square when (b, c) = (2, 1), (3, 2), (4, 3), (4, 4) (5, 4). This gives five possible triples.

Case 2: $a = 2, 3≤b≤5$

Discriminant is $b^2 - 8c$ .

The discriminant is a perfect square when (b, c) = (3, 1), (4, 2), (5, 2), (5, 3), giving us four possible triples.

Case 3: $a = 3, b = 4, 5$

Discriminant is $b^2 - 12c$ .

b must equal either 4 or 5.

The discriminant is a perfect square when (b, c) = (4, 1), (5, 2), giving us two possible triples.

Case: $a = 4, b = 4, 5$

Discriminant is $b^2 - 16c$ .

b must equal either 4 or 5.

The discriminant is a perfect square when (b, c) = (4, 1), (5, 1), giving us two possible triples.

Case: $a = 5, b = 5$

Discriminant is $b^2 - 20c$ .

Since b must equal 5, there are no cases where the discriminant is a perfect square. Thus there are no possible triples in this case.

Adding up all our triples, we have $5 + 4 + 2 + 2 = 13$ possible ordered triples (a, b, c) such that $ax^2 + bx + c$ has a rational solution.

From the Quadratic Formula, \frac{-b \pm \sqrt{b^2-4ac}}{2a}, we know that our roots will not be rational if we have a square root term, so our discriminant, b^2-4ac, must be a perfect square. Since b is limited to at most 5 and a and c are both limited to at least 1, the maximum value of the discriminant is 21, so we only need to consider the perfect squares up to 16. We then consider each case:

case 1: b^2-4ac=0

b^2 must be a multiple of 4, so b can be either 2 or 4.

If b=2, then 4ac=4, and a=c=1. This gives us our first triple; (1, 2, 1).

If b=4, then 4ac=16 \rightarrow ac=4 \rightarrow (a, b, c) = (1, 4, 4), (2, 4, 2), or (4, 4, 1)

case 2: b^2-4ac=1

b can be either 3 or 5.

If b=5, 4ac=24 \rightarrow ac=6 \rightarrow (a, b, c) = (2, 5, 3), (3, 5, 2).

If b=3, 4ac=8 \rightarrow ac=2 \rightarrow (a, b, c) = (2, 3, 1), (1, 3, 2).

case 3: b^2-4ac=4

b must be 4. If b=4, 4ac=12 \rightarrow ac=3 \rightarrow (a, b, c) = (1, 4, 3), (3, 4, 1).

case 4: b^2-4ac=9

b must be 5. If b=5, 4ac=16 \rightarrow ac=4 \rightarrow (a, b, c) = (1, 5, 4), (2, 5, 2), (4, 5, 1).

considering b^2-4ac=16 we find there are no triples (a, b, c) given the constraints, thus we have found all of our triples. Counting the number of triples we arrive at 13 as our answer.

The solutions of $ax^2 + bx + c$ are $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
To make $ax^2+bx+c$ has a rational solution, $\sqrt{b^2-4ac}$ must be an integer or $b^2-4ac$ must be a perfect square
We divide it into 5 cases
Case 1 : $b=1$
$b^2-4ac$ is a perfect square
$1-4ac$ is a perfect square
Since $1 \leq a,b,c \leq 5$ , then $1-4ac \leq 1-4(1)(1) < 0$
Therefore, $1-4ac$ cannot be a perfect square
Case 2: $b=2$
$b^2-4ac$ is a perfect square
$4-4ac$ is a perfect square
The greatest possible value of $4-4ac$ is $4-4(1)(1)=0$
To make $4-4ac$ a perfect square, $4-4ac$ must be equal to $0$
So, $a=c=1$
In case 2, there is $1$ triple positive integer $(a,b,c)$
Case 3: $b=3$
$b^2-4ac$ is a perfect square
$9-4ac$ is a perfect square
The greatest possible value of $9-4ac$ is $9-4(1)(1)=5$
To make $9-4ac$ a perfect square, $9-4ac$ must be equal to $0,1,$ or $4$
Since $9-4ac$ is an odd number, then $9-4ac$ must be $1$ or $ac=2$ The possible values of $(a,b,c)$ in this case are ${(1,3,2), (2,3,1)}$
In case 3, there are $2$ triple positive integers $(a,b,c)$
Case 4: $b=4$
$b^2-4ac$ is a perfect square
$16-4ac$ is a perfect square
The greatest possible value of $16-4ac$ is $16-4(1)(1)=12$
To make $16-4ac$ a perfect square, $16-4ac$ must be equal to $0,1,4$ or $9$
Since $16-4ac$ is an even number, then $16-4ac$ must be $0$ or $4$
When $16-4ac=0$ , $ac=4$
The possible values of $(a,b,c)$ are ${(1,4,4), (2,4,2), (4,4,1)}$
When $16-4ac=4$ , $ac=3$
The possible values of $(a,b,c)$ are ${(1,4,3), (3,4,1)}$
In case 4, there are $2+3=5$ triple positive integers $(a,b,c)$
Case 5: $b=5$
$b^2-4ac$ is a perfect square
$25-4ac$ is a perfect square
The greatest possible value of $25-4ac$ is $25-4(1)(1)=21$
To make $25-4ac$ a perfect square, $25-4ac$ must be equal to $0,1,4,9$ or $16$
Since $25-4ac$ is an odd number, then $25-4ac$ must be $1$ or $9$
When $25-4ac=1$ , $ac=6$
The possible values of $(a,b,c)$ are ${(2,5,3), (3,5,2)}$
When $25-4ac=9$ , $ac=4$
The possible values of $(a,b,c)$ are ${(1,5,4), (4,5,1), (2,5,2)}$
In case 5, there are $2+3=5$ triple positive integers $(a,b,c)$
Therefore, from all cases, there are $0+1+2+5+5 = 13$ triple positive integers $(a,b,c)$

Assume that the equation $ax^2+bx+c=0$ has a rational number, namely, $\frac{p}{q}$ ( $p,q \in \mathbb{Z^+}$ ). which means $ap^2+bpq+cq^2=0 \Leftrightarrow (2ap+bq)^2=q^2(b^2-4ac)$ (\Rightarrow b^2-4ac ) is a square of a rational number .Since $b^2-4ac \in \mathbb{ Z}$ , we imply that: $b^2-4ac$ is a square of a integer. In short: $b^2-4ac=x^2$ for some positive integers x. As $b$ ranges from 1 to 5 .We can find number of triples and our desired number is 13

We want $b^2-4ac$ to be an integer, and simple casework finishes the problem.

since there are only 4 values of b {2 3 4 5} are possible, calculate total no. of possible solution for each value of b. we calculate discriminant b b - 4 a*c (for rational solution, it should be perfect square)

for b=2 , a=1 and c=1 only combination

for b=3 , (a=2 and c=1) or (a=1 and c=2) two ways

for b=4 , (4,1), (1,4), (3,1), (1,3), (2,2) five ways

for b=5 , (4,1), (1,4), (2,2), (2,3), (3,2) five ways

Hence total 13 combinations are possible, such that ax2+bx+c=0 has a rational solution.

b^2 can assume only the values 4,9,16,25, while 4ac has to be chosen from 4, 8,12,16, 24 to ensure b^2 - 4ac is a non-negative perfect square. So we have the following possibilities (i) (4,4), (9,8), (16,16),(16,12),(25,16),(25,24) for (b^2, 4ac) (ii) (2,1),(3,2),(4,4),(4,3),(5,4),(5,6) resp for (b, ac) These give rise to the following 13 values for the triplet (b,a,c): (2,1,1); (3,1,2),(3,2,1); (4,1,4),(4,4,1),(4,2,2); (4,1,3),(4,3,1); (5,1,4),(5,4,1),(5,2,2),(5,2,3),(5,3,2); Required triplets (a,b,c) [got by switching a,b in each (a,b,c) above] are also 13 in number

To find-number of ordered triplet a,b,c such that $ax^2+bx+c$ has rational solution. Given- a,b,c are integers and lie between 1 and 5

now the roots of given equation are- $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ now to have rational solution the discriminant of the equation must be a perfect square as a,b,c are integers.

suppose D= $b^2- 4*a*c$ is K^2. also k will always be less than b as (a,b,c are integers lying between 1 and 5)

Case 1 b=1 (no solutions in integers a,b,c)

Case 2 b=2 then only 1 solution possible as- $b^2-4*a*c=k^2$ hence possible values of k^2 are 1 and 0 2.1) $b^2-4*a*c=1^2 4-4*a*c=1 3=4*a*c$ but this is not possible as and c are integers. 2.2) $b^2-4*a*c=0^2$ $4-4*a*c=0$ ac=1 hence a and c are 1 and 1 respectively. hence only one solution in case 2.

similarily case 3 has 2 solutions and case 4 and 5 both have5 solutions each.

Hence adding all we get 13 solutions to this problem.and these are ordered triplet as in any solution a,b,c 2,3,1 is different from 1,3,2.

Enumerate

The equation $ax^2+bx+c=0$ will have rational solutions if its determinant is a perfect square, that is, $b^2-4ac$ is a perfect square. Now the maximum value of $b^2-4ac$ is $5^2-4(1)(1)=21$ , so possible values of $b^2-4ac$ are $0, 1, 4, 9$ and $16$ .

If $b=1$ , then $4ac = 1, 0, -3, -8$ or $-15$ . None of them are possible.

If $b=2$ , then $4ac = 4, 3, 0, -1$ or $-12$ . Then $ac = 1$ is the only one that is possible. We have $a = 1$ and $c = 1$ , giving us 1 triple.

If $b=3$ , then $4ac = 9, 8, 5, 0$ or $-7$ . Then $ac = 2$ is the only one that is possible. We have $a = 2$ and $c = 1$ , or $a = 1$ and $c = 2$ , giving us 2 triples.

If $b=4$ , then $4ac = 16, 15, 12, 7$ or $0$ . Then $ac = 4$ or $ac = 3$ . Now $4 = 1 \times 4 = 2 \times 2 = 4 \times 1$ and $3 = 1 \times 3 = 3 \times 1$ , so we have 5 triples.

If $b=5$ , then $4ac = 25, 24, 21, 16$ or $9$ . Then $ac = 6$ or $ac = 4$ . We have $6 = 1 \times 6 = 2 \times 3 = 3 \times 2 = 6 \times 1$ and $4 = 1 \times 4 = 2 \times 2 = 4 \times 1$ , but neither $a$ nor $c$ can have a value of $6$ . So we have 5 triples only.

$1 + 2 + 5 + 5 = 13$ triples.

In order to have a rational root the delta of the equation has to be a perfect square. So the problem is equal to find the possible solutions to b^2-4ac=k^2, where k is an integer. Now we see the different cases with b=1,2,3,4,5. If b=1 we have no solution. If b=2 we have the only solution (1,2,1). If b=3 we have (2,3,1) and (1,3,2) as solutions. If b=4 we have (2,4,2),(4,4,1),(1,4,4),(3,4,1),(1,4,3) as solutions. If b=5 we have (3,5,2),(2,5,3),(4,5,1),(1,5,4),(2,5,2) as solutions. In this way we see that the different triples are 13.

For the quadratic expression $ax^2+bx+c$ to have a rational solution, its discriminant must be the square of a rational number, so $b^2-4ac=D^2$ , where $D$ is rational. Furthermore, we note that since $a, b$ and $c$ are all integers, $D$ must be an integer too. From here, we can simply list out possibilities for $b$ and find corresponding values of $a$ and $c$ .

When $b=1, D^2=1-4ac<0$ , a contradiction since $D$ must be rational, so there are no triples.

When $b=2, D^2=4-4ac\geq0$ , and there is only the solution $(1,2,1)$ .

When $b=3, D^2=9-4ac\geq0$ , and we can get $(1,3,2), (2,3,1)$ .

When $b=4, D^2=16-4ac\geq0$ simplifies to $4-ac\geq0$ , so we have $(2,4,2), (1,4,3), (3,4,1), (1,4,4), (4,4,1)$ .

When $b=5, D^2=25-4ac\geq0$ and the solutions are $(2,5,2), (1,5,4), (4,5,1), (2,5,3), (3,5,2)$ .

Adding the number of solutions for each possibility, we have a total of $1+2+5+5=13$ ordered triples.

Claim: If $\sqrt{N}$ is rational then $N$ must be a perfect square. (This was stated in the blog post, and proved in Rational Numbers II .

Proof: Suppose $\sqrt{N} = \frac{a}{b}$ , for some integers $a$ and $b$ where $a$ and $b$ are coprime. So, we have $b^2N = a^2$ . If a prime, $p$ , divides $b$ then it must divide $b^2N$ thus it must divide $a^2$ which means it divides $a$ . Since $a$ and $b$ are coprime, no such $p$ exists and this implies that $b=1$ . Therefore $N = a^2$ is a perfect square.

Since $a, b$ and $c$ are positive integers, $ax^2 + bx+c$ has a rational solution if, and only if, the discriminant $b^2 - 4ac$ is a perfect square (as shown in the claim). We proceed by fixing $b$ and considering the pairs $(a, c)$ . Observe that $\sqrt{b^2-4ac}$ will have the same parity as $b$ , and is strictly smaller than $b$ since $ac$ is positive.

Case 1: $b=5$ . We have $b^2-4ac = 1, 9$ , which gives $ac = 6, 4$ . This has solutions $(a,c) = (2,3), (3,2), (1,4), (2,2), (4,1)$ for a total of 5.

Case 2: $b=4$ . We have $b^2-4ac = 0, 4$ , which gives $ac=4, 3$ . This has solutions $(a,c) = (1,4), (2,2), (4,1), (1,3), (3,1)$ for a total of 5.

Case 3: $b=3$ . We have $b^2 -4ac = 1$ , which gives $ac= 2$ . This has solutions $(a,c)=(1,2), (2,1)$ for a total of 2.

Case 4. $b=2$ . We have $b^2 -4ac = 0$ , which gives $ac=1$ . This has solutions $(a,c) = (1,1)$ for a total of 1.

Case 5. $b=1$ . We have no possible values for $b^2-4ac$ .

Hence, the total number of solutions is $5+5+2+1+0=13$ .