Below, we present a problem from the 12/17 Algebra and Number Theory set, along with 3 student submitted solutions (none of them have been edited). You may vote up or down for the solutions that you think should be featured / or are wrong.

Problem: Cute cubic with integer roots Suppose that the polynomial \(p(x)=x^3−ax^2+bx−c\) has 3 positive integer roots and that \(4a+2b+c=1741\). Determine the value of \(a\).

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

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

Note: I will not reveal who wrote which solution in this blog post. The chosen featured solution will (as always) be credited to the originator.

Remarks about solutions -

Solution A is false due to the statement that "k+4=33" is the only possibility. Due to the cyclic nature of the roots, there is no guarantee that we picked \( \{x, y\} = \{ 1, 9 \} \). This was a common mistake in most solutions submitted.

Solution B is not completely true, because he left out using the fact that the roots are positive. For example, we also have \(-1749 = -1 \times -1 \times -1749\), which could have been used as a possible factorization.

I agree with the Brilliant community that Solution C submitted by Abhishek K. is the best. It explains where we used the fact that the roots are positive integers, to approach the problem.

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## Comments

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TopNewestSolution C - Observation, Viete's Theorem

Let \(\alpha, \beta, \gamma\) be the roots of \(p(x)\). Thus \(p(x)=(x-\alpha)(x-\beta)(x-\gamma)\). We observe that \(p(-2)=-8-(4a+2b+c)=-8-1741=-1749\). We also note that \(3\times 11\times 53=1749\). Thus we get \((2+\alpha)(2+\beta)(2+\gamma)=3\times 11\times 53\). Since \(\alpha,\beta,\gamma >0\) and each one is an integer we must have \(\{\alpha,\beta,\gamma\}=\{3-2,11-2,53-2\}=\{1,9,51\}\). By Viete's theorem we also have \(\alpha+\beta+\gamma=a\), thus \(a=1+9+51=61\).

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Solution A - Simplify and Factorise

Let the \(3\) positive integer roots be \(x\),\(y\) and \(z\) Given , \(4(x+y+z) + 2(xy+yz+zx) + xyz =1741\) which simplifies to , \(z(4+(2x+2y+xy))+2(2x+2y+xy)=1741\) Taking \(2x+2y+xy=k\) \(z(4+k)+2k=1741\) \(z= \frac{1749}{4+k} - 2\) , now since \(z\) is an integer \(4+k\) must divide \(1749\) and we have \(1749=3*11*53\). So we are only left with some few cases like \(k+4 = 3\) or \(11\) or \(53\) and so on..after checking we find that \(k+4=33\) is the only possibility i.e \(2x+2y+xy=29\) which gives \(x=1\) ,\( y=9 \) and \(z=53-2=51\) Therefore we have \(a=x+y+z = 1+9+51=61\)

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Solution B - Roots of Polynomials

Plugging in \(x = -2\), \(P( -2 ) = - 8 - (4a + 2b + c) = - 8 - 1741 = - 1749 =(- 3)(- 11)(- 53)\) \(= (- 2 - 1)(- 2 - 9)(- 2 - 51)\) \(P( x ) = (x - 1)(x - 9)(x - 51)\) Therefore, by Vieta's Formula, we get \(a = 1 + 9 + 51 = 61\).

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ahh, this was the only problem i didn't know this week in algebra and number theory. i made the connection with f(-2), but failed to go any further.

anyways, solution A has my vote

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Solution B and C are similar

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