# [Calvin] Which solution will you feature (11)?

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Below, we present a problem from the 2/25 Algebra and Number Theory set, along with 3 student submitted solution. You may vote up for the solutions that you think should be featured, and should vote down for those solutions that you think are wrong.

Integrally rational How many ordered triples of positive integers $(a, b, c)$ with $1 \leq a, b, c \leq 5$ are there such that $a x^2 + b x + c = 0$ has a rational solution?

You may try the problem by clicking on the above link.

All solutions may have LaTeX edits to make the math appear properly. The exposition is presented as is, and has not been edited.

If you think that all these solutions are essentially the same, read VERY carefully. They differ in one important area. Most submitted solutions were not marked correct.

There are 2 parts to this question. The first part involves arguing that a solution occurs if and only if $b^2 - 4ac$ is a perfect square, and the second part involves actually counting the number of possibilities. Most students did not do the first part well.

Solution A - This solution didn't do the second part. If you look at his logical implications, he only showed that "If $ax^2 + bx +c = 0$ has a rational number", then "$b^2 - 4ac$ is the square of an integer". He then proceeds to state that the number of cases where $b^2 - 4ac = x^2$ is 13. This doesn't answer the original question, and merely shows that the answer is at most 13.

Solution B - This logical deductions in this solution does not hold. It is not true that " (A) To make $ax^2 + bx + c$ has a rational solution", "(B) $\sqrt{b^2-4ac}$ must be an integer", "(C) or $b^2 - 4ac$ must be a perfect square. In fact, all that (A) implies is "$\sqrt{b^2 - 4ac}$ must be a rational number". We then need to show that "since $b^2 - 4ac$ is an integer, hence $b^2 - 4ac$ is a perfect square". Likewise, this does not explain why those 13 cases satisfy the original conditions.

Solution C - This statement of " one of these numbers is rational if and only if $b^2 - 4ac = k^2$" is the only correct solution. It has an error in Case 5 as pointed out by Bob. This solution is presented by Sreejato. Note by Calvin Lin
8 years, 1 month ago

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Solution B - 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 $(a,b,c)$

Staff - 8 years, 1 month ago

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Remarks have been added.

Staff - 8 years, 1 month ago

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Solution C - We know that the roots of $ax^2 + bx + c$ are $(-b+\sqrt{b^2-4ac})/2a$ and $(-b- \sqrt{b^2-4ac})/2a$. 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 4 solutions over the integers. So case 5 gives 7 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.

Staff - 8 years, 1 month ago

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Wait a second....adding the numbers from the casework gets 0+1+2+5+7=13? Last time I checked, it didn't. Either this is a typo, or it's a flaw that the author just decided to glance over. The key is that in case 5, there are only 5 solutions. If ac=6, a and c must be 2 or 3 and no other numbers; niether can be 6 by the bounds mentioned in the problem. Proofreading is a crucial step to a solid proof.

- 8 years, 1 month ago

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That's correct, you spotted 1 mistake with this proof. Like you said, $ac=6$ doesn't have 4 solutions, because $1 \times 6, 6 \times 1$ is not valid.

Staff - 8 years, 1 month ago

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it isnt right... the qeu asks for rational solutions and not for integral solution... so it is not necessary for b^2−4ac to be a perfect square

- 8 years, 1 month ago

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Correct, b^2-4ac doesn't have to be a perfect square, it just needs to be the square of a rational number. However, since a, b, and c are all integers defined by the problem, b^2-4ac must also be an integer. And so if one looks for rational solutions, b^2-4ac must be a square.

- 8 years, 1 month ago

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yup agreed.

- 8 years, 1 month ago

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Solution A - 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.

Staff - 8 years, 1 month ago

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If I did not know how to do the problem, after reading this I would still not know how to do the problem.

- 8 years, 1 month ago

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