Difference of Two CubesDate: 11/18/96 at 20:48:48 From: James Fletcher Subject: Difference of two cubes Dear Dr. Math I am wondering if there is an algebraic method to solve the following problem: The difference of two cubes is 56,765. What two positive integers satisfy this condition? Thank you very much, James Fletcher Date: 11/21/96 at 19:18:37 From: Doctor Pete Subject: Re: Difference of two cubes I'm not sure what you mean by "algebraic," but you can use some algebra to obtain some additional conditions on the solution, if one exists. In particular, note that the difference of two cubes can be factored: a^3 - b^3 = (a-b)(a^2 + ab + b^2) So let's assume that a > b. If a^3 - b^3 = 56765, then if an integer solution exists, a-b and a^2 + ab + b^2 must be factors of 56765. Indeed, we see that 56765 = 5(11353), and this latter factor is prime (a bit of arithmetic verifies this). Therefore, we have: a-b = 1, a^2 + ab + b^2 = 56765, or a-b = 5, a^2 + ab + b^2 = 11353 (note that a-b < a^2 + ab + b^2, because a^2 + ab + b^2 = (a-b)^2 + 3ab.) If we assume the first case, then a = b+1, so (b+1)^3 - b^3 = 3b^2 + 3b + 1 = 56765 or 3b^2 + 3b = 56764. But the lefthand side has 3 as a factor, whereas the righthand side is not divisible by 3 (the sum of the digits is 38). Hence this leads to an impossibility, so this case is not true. In the second case, we have a = b+5, and thus (b+5)^3 - b^3 = 15b^2 + 75b + 125 = 56765 or b^2 + 5b - 3776 = 0. At this point, we may use the quadratic equation to solve for b. We obtain: b = (-5 + Sqrt[25+4(3776)])/2 = (-5+123)/2 = 59, (I have omitted the negative root, since the question asks for only positive integer solutions). Thus b = 59, a = 59+5 = 64, and indeed, we can verify that 64^3 - 59^3 = 262144 - 205379 = 56765. The crucial point here is to notice that the difference of cubes can be factored, and you can try to match up factorizations of the given integer with these. It doesn't always work, and sometimes, it works in several cases (unlike here, where it works in exactly one case). So it's important to try them out. Furthermore, notice that the factorization essentially gives us two equations in two unknowns, and that the difference of cubes here allows the degree of these equations to be small enough to work with. If one were given something like the difference of fifth powers, then we would not necessarily be able to complete the last step, which relies on the fact that we get a quadratic equation which is easily solved. Of course, we get a solution, but it doesn't strike me as being a method that guarantees an answer; this is why I hesitate to call it a *definitive* solution. All I did was use some additional facts about differences of cubes to obtain extra conditions on a and b so that they could be determined - as I mentioned above, this doesn't always work. Hope this helps, -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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