[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) (a, b, c) with 1a,b,c5 1 \leq a, b, c \leq 5 are there such that ax2+bx+c=0 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 b24ac 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 ax2+bx+c=0 ax^2 + bx +c = 0 has a rational number", then "b24acb^2 - 4ac is the square of an integer". He then proceeds to state that the number of cases where b24ac=x2b^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 ax2+bx+cax^2 + bx + c has a rational solution", "(B) b24ac \sqrt{b^2-4ac} must be an integer", "(C) or b24acb^2 - 4ac must be a perfect square. In fact, all that (A) implies is "b24ac \sqrt{b^2 - 4ac} must be a rational number". We then need to show that "since b24acb^2 - 4ac is an integer, hence b24acb^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 b24ac=k2 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
7 years, 2 months ago

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7 votes

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Solution B - The solutions of ax2+bx+c ax^2 + bx + c are b±b24ac2a \frac{-b \pm \sqrt{b^2-4ac}}{2a} To make ax2+bx+c ax^2+bx+c has a rational solution, b24ac \sqrt{b^2-4ac} must be an integer or b24ac b^2-4ac must be a perfect square We divide it into 5 cases

Case 1 : b=1 b=1 b24ac b^2-4ac is a perfect square 14ac 1-4ac is a perfect square Since 1a,b,c5 1 \leq a,b,c \leq 5 , then 14ac14(1)(1)<0 1-4ac \leq 1-4(1)(1) < 0 Therefore, 14ac 1-4ac cannot be a perfect square

Case 2: b=2 b=2 b24ac b^2-4ac is a perfect square 44ac 4-4ac is a perfect square The greatest possible value of 44ac 4-4ac is 44(1)(1)=0 4-4(1)(1)=0 To make 44ac 4-4ac a perfect square, 44ac 4-4ac must be equal to 0 0 So, a=c=1 a=c=1 In case 2, there is 1 1 triple positive integer (a,b,c) (a,b,c)

Case 3: b=3 b=3 b24ac b^2-4ac is a perfect square 94ac 9-4ac is a perfect square The greatest possible value of 94ac 9-4ac is 94(1)(1)=5 9-4(1)(1)=5 To make 94ac 9-4ac a perfect square, 94ac 9-4ac must be equal to 0,1, 0,1, or 4 4 Since 94ac 9-4ac is an odd number, then 94ac 9-4ac must be 1 1 or ac=2 ac=2 The possible values of (a,b,c) (a,b,c) in this case are (1,3,2),(2,3,1) {(1,3,2), (2,3,1)} In case 3, there are 2 2 triple positive integers (a,b,c) (a,b,c)

Case 4: b=4 b=4 b24ac b^2-4ac is a perfect square 164ac 16-4ac is a perfect square The greatest possible value of 164ac 16-4ac is 164(1)(1)=12 16-4(1)(1)=12 To make 164ac 16-4ac a perfect square, 164ac 16-4ac must be equal to 0,1,4 0,1,4 or 9 9 Since 164ac 16-4ac is an even number, then 164ac 16-4ac must be 0 0 or 4 4 When 164ac=0 16-4ac=0 , ac=4 ac=4 The possible values of (a,b,c) (a,b,c) are (1,4,4),(2,4,2),(4,4,1) {(1,4,4), (2,4,2), (4,4,1)} When 164ac=416-4ac=4 , ac=3 ac=3 The possible values of (a,b,c) (a,b,c) are (1,4,3),(3,4,1) {(1,4,3), (3,4,1)} In case 4, there are 2+3=5 2+3=5 triple positive integers (a,b,c) (a,b,c)

Case 5: b=5 b=5 b24ac b^2-4ac is a perfect square 254ac 25-4ac is a perfect square The greatest possible value of 254ac 25-4ac is 254(1)(1)=21 25-4(1)(1)=21 To make 254ac 25-4ac a perfect square, 254ac 25-4ac must be equal to 0,1,4,9 0,1,4,9 or 16 16 Since 254ac 25-4ac is an odd number, then 254ac 25-4ac must be 1 1 or 9 9 When 254ac=1 25-4ac=1 , ac=6 ac=6 The possible values of (a,b,c) (a,b,c) are (2,5,3),(3,5,2) {(2,5,3), (3,5,2)} When 254ac=925-4ac=9 , ac=4 ac=4 The possible values of (a,b,c) (a,b,c) are (1,5,4),(4,5,1),(2,5,2) {(1,5,4), (4,5,1), (2,5,2)} In case 5, there are 2+3=5 2+3=5 triple positive integers (a,b,c) (a,b,c)

Therefore, from all cases, there are 0+1+2+5+5=13 0+1+2+5+5 = 13 triple positive integers (a,b,c) (a,b,c)

Calvin Lin Staff - 7 years, 2 months ago

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Remarks have been added.

Calvin Lin Staff - 7 years, 2 months ago

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Solution C - We know that the roots of ax2+bx+cax^2 + bx + c are (b+b24ac)/2a(-b+\sqrt{b^2-4ac})/2a and (bb24ac)/2a(-b- \sqrt{b^2-4ac})/2a. Now it can easily be proved that one of these numbers is rational if and only if b24ac=k2b^2-4ac= k^2 for some integer kk. The given problem can now be divided into 5 cases:- b=1,b=2,b=3,b=4b=1, b=2, b=3, b=4, and b=5b=5.

Case 1 b=1b=1
The equation now becomes 14ac=k21-4ac= k^2. This obviously has no solution over the integers since 14ac1411=3<01-4ac\leq 1-4*1*1=-3<0, but the square of a real number has to be non-negative.

Case 2 b=2b=2
The equation now becomes 44ac=k24-4ac= k^2 , or 4(1ac)=k24(1-ac)= k^2. Since 4 is a perfect square and so is k2k^2, 1ac1-ac must also be a perfect square. Since 1ac11-ac\leq 1 (a and c being positive integers) and k20k^2\geq 0, the only possible value for 1ac1-ac is 0, which implies ac=1ac=1. This has one solution over the integers, which is (a,c)=1(a, c)= 1. So case 2 gives one solution over the integers.

Case 3 b=3b=3
The equation now becomes 94ac=k29-4ac=k^2. Since 91(mod4)9\equiv1\pmod{4} and 4ac0(mod4)4ac\equiv0\pmod{4} so k21(mod4)k^2\equiv1\pmod{4}. Also 94ac9411=59-4ac\leq 9-4*1*1= 5. The only perfect square less than 5 and congruent to 1mod 4 is 1. So 94ac=19-4ac= 1, or ac=2ac= 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=4b=4
The equation now becomes 164ac=k216-4ac= k^2, or 4(4ac)=k24(4-ac)= k^2. Since 4 is a perfect square and so is k2k^2, 4ac4-ac must also be a perfect square. This gives 2 possible values for ac:- ac=4ac=4 and ac=3ac=3. The case ac=4ac=4 has 3 solutions over the integers, and the case ac=3ac=3 has 2 solutions over the integers. So case 4 gives 5 solutions over the integers.

Case 5 b=5b=5
The equation now becomes 254ac=k225-4ac= k^2. Since 251(mod4)25\equiv1\pmod{4} and 4ac1(mod4)4ac\equiv1\pmod{4}, k21(mod4)k^2\equiv1\pmod{4}. Also k2<25k^2<25. The only perfect squares less than 25 and congruent to 1 mod 4 are 9 and 1. The case k2=9k^2=9 gives ac=4ac= 4, which has 3 solutions over the integers. The case k2=1k^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 bb is different in different cases. So adding we get that the number of solutions is 13.

Calvin Lin Staff - 7 years, 2 months 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.

Bob Krueger - 7 years, 2 months ago

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

Calvin Lin Staff - 7 years, 2 months 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

Devvrit Khatri - 7 years, 2 months 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.

Bob Krueger - 7 years, 2 months ago

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yup agreed.

Devvrit Khatri - 7 years, 2 months ago

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Solution A - Assume that the equation ax2+bx+c=0 ax^2+bx+c=0 has a rational number, namely, pq \frac{p}{q} ( p,qZ+ p,q \in \mathbb{Z^+} ), which means ap2+bpq+cq2=0(2ap+bq)2=q2(b24ac) ap^2+bpq+cq^2=0 \Leftrightarrow (2ap+bq)^2=q^2(b^2-4ac) b24ac\Rightarrow b^2-4ac is a square of a rational number .Since b24acZ b^2-4ac \in \mathbb{ Z} , we imply that: b24ac b^2-4ac is a square of a integer. In short: b24ac=x2 b^2-4ac=x^2 for some positive integers x. As b b ranges from 1 to 5 .We can find number of triples and our desired number is 13.

Calvin Lin Staff - 7 years, 2 months 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.

Bob Krueger - 7 years, 2 months ago

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