Cute cubic with integer roots

This problem has an algebraic flavour too IMO. For instance, once we see that we are given a relationship regarding the roots of an equation, we should consider vieta's formula . However, directly spamming Vieta's lead us nowhere since we need information of the roots and let them be $\alpha, \beta$ and $\gamma$ .

Another key motivation here is that we need to link $4a+2b+c = 1741$ to the polynomial $p(x)$ . One nice way to do so, is to substitute into the polynomial some key values . For instance, we would like a term $4a$ to be generated so we might consider subbing in $x = ±2$ (so that $x^2 = 4$ ).

Combining the above two critical observations, we can realise, intuitively, in order for the $+, -$ signs in $p(x)$ to match with that of $4a + 2b + c$ , we will like to consider $p(-2)$ . Also, in order to bring in the roots, we write $p(x)$ as $(x - \alpha)(x- \beta)( x - \gamma)$ . Now, we have: (using the expression given)

$p(-2) = (-2)^3 - a(-2)^2 + b(-2) - c = 8 - (4a+2b+c) = 8 - 1741 = -1749$ Aha! We eliminated the variable $x$ from the expression. We also have, writing $p(-2)$ in another way,

$p(-2) = ( -2 - \alpha)(-2 - \beta)(-2 - \gamma) = -(2+ \alpha)(2+ \beta)(2+ \gamma)$

Equating and simplifying, we have:

$(2+ \alpha)(2+ \beta)(2+ \gamma) = 1749$

At this point, intuitively, we would like to factorise $1749$ (to split it into $3$ terms) which comes out as $3\times 11\times 53$ . So WLOG, $\alpha = 3-2 = 1$ , $\beta = 11-2 = 9$ and $\gamma = 53 - 2 = 51$ .

Turning finally to the first idea, Vieta gives $a = \alpha + \beta + \gamma = 51 + 9 + 1 = \fbox{61}$ , the desired answer.

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 $a=1+9+51=61$ .

We substitute x = -2 to get -8-4a-2b-c = f(-2). Thus f(-2) = -1749 = (-3) (-11) (-53). If we let f have roots i,j,k, then (-2-i)(-2-j)(-2-k) =(-3) (-11) (-53). Since i, j, and k are positive integers, i,j,k must be 1,9,and 51 in some order, so a = 1+9+51 = 61

Let x=-2.

$p(-2) = (-2)^3 - a(-2)^2 + b(-2) - c = -8 -(4a+2b+c) = -1749$

Since the polynomial has 3 positive integer roots and is monic (has a leading coefficient of 1),

$p(x)=(x-r_1)(x-r_2)(x-r_3)$ , where $r_1$ , $r_2$ and $r_3$ are the positive roots.

This implies that $-1749=(-2-r_1)(-2-r_2)(-2-r_3)$ .

Multiplying both sides by $-1$ , we get $1749=(2+r_1)(2+r_2)(2+r_3)$ .

Note that $1749=3 \times 11 \times 53$ .

Since $2+r_1$ , $2+r_2$ and $2+r_3$ are positive integers greater than 3 (due to the fact that $r_1$ , $r_2$ and $r_3$ are positive), we know that they have to correspond to 3, 11 and 53 in some order.

Suppose one of them is not 3, 11 or 53. Without loss of generality, let it be $2+r_1$ that is not one of these. Then it would need to be an integer. Since it is not 3, 11 or 53, we know it would have to be some other factor of 1749.

Suppose it is not a factor of 1749. Consider the product $(2+r_2)(2+r_3)$ . Note that it would not be an integer. We know that at least one of $2+r_2$ and $2+r_3$ are not integers. Else if both $2+r_2$ and $2+r_3$ are integers, the product will be an integer and thus contradicting the fact that $(2+r_2)(2+r_3)$ is not an integer.

Thus we have proven it is a factor of 1749. Since it is not 3, 11 or 53, it would have to be one of the composite factors of 1749, since obviously, it cannot be 1, else from $2+r_1=1$ we get $r_1=-1$ , contradicting the fact $r_1$ is a positive integer.

Since 1749 is the product of 3 prime numbers, we will have the product of the other 2 factors either a prime number or 1.

Suppose the product of the other factors is not a prime number. Then it must have at least 2 prime factors in its prime factorization. Multiplying it by $2+r_1$ , we get a number with at least 4 prime factors in its prime factorization, contradicting the fact that there are 3 prime numbers in the prime factorization of 1749.

Thus we know $(2+r_2)(2+r_3)$ is prime. Since each of these factors of this prime number are positive integers, we are sure that one is that prime and the other is 1, since they are the only 2 factors of the prime.

WLOG, let $2+r_2=1$ . Then $r_2=-1$ , contradicting the fact that $r_2$ is positive.

Thus, we know that $(2+r_2)(2+r_3)$ has to be composite. But this contradicts the fact $2+r_1$ is composite.

Thus we know $2+r_1$ is prime.

By symmetry of the argument, $2+r_2$ and $2+r_3$ are prime.

Thus $r_1+r_2+r_3=(3-2)+(11-2)+(53-2)=61$

By Vieta's formulas, we get

$a=-(-a)=r_1+r_2+r_3=61$ .

Let p(x)=x^{3}-a(x^{2})+bx-c = (x-i)(x-j)(x-k) with i,j ,k being integers \geq 1. p(-2)=-8-(4a+2b+c)=-(2+i)(2+j)(2+k) 1741=4a+2b+c This gives (2+i)(2+j)(2+k)=1749=3.11.53 By the restriction on i, j, k to be positive integers, in above equation the factors on the left match with those on the right upto permutations. i=1, j=9, k=51 a=sum of roots= 61

Notice that $-p(-2) = 8 + 4a + 2b + c = 1749$ . Furthermore, we can factor the polynomial $p(x)$ using its roots $A, B, C$ as $p(x) = (x - A)(x - B)(x-C).$ Therefore $(A+2)(B+2)(C+2) = 1749$ . The factorization of $1749$ is $3 \cdot 11 \cdot 53$ and because the roots are integral we must have $A = 1, B = 9, C = 51$ . Finally, $a = A + B + C = 1 + 9 + 51 = {\bf 61}$ .

Assume that three roots are r1,r2,and r3, respectively. Notice that r1+r2+r3=a, r1r2+r1r3+r2r3=b, and r1r2r3=c since every cubic equation can be written in the form (x-r1)(x-r2)(x-r3). It follows that (r1+2)(r2+2)(r3+2)=4a+2b+c+8=1749=3 11 53. The values of three roots are 1, 9, and 51, respectively. Therefore, the value of a is 61.

First of all, we see that all the roots of this polynomial are integers, since it has degree 3. Now, the idea that one should always have is factorising -- but what to factorise?

In our help comes the second relation, $4a+2b+c=1741$ ; it looks like a second degree polynomial evalued in $2$ - but we actually have something similar to it, which is the $-ax^2+bx-c$ part of $p(x)$ ; those " $-$ "s are saying: "take $-2$ "! So we have $p(-2)=-8-4a-2b-c=-8-1741=-1749$ Here we are! We have "polynomial=integer" and we are mostly done: by the fundamental Theorem of Algebra, we write $p(x)=(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$ where $\lambda_1,\lambda_2,\lambda_3$ are the three integer roots of $p(x)$ .

Taking $x=-2$ we get $(-2-\lambda_1)(-2-\lambda_2)(-2-\lambda_3)=-1749$ Eliminating the $-$ s and factorising $1749=3\cdot11\cdot53$ we finally obtain $(2+\lambda_1)(2+\lambda_2)(2+\lambda_3)=3\cdot11\cdot53$

Since the roots are positive, each factor on the Left Hand Side is bigger than $2$ , so it can't be $1$ . Hence it follows, assuming without loss of generality $\lambda_1\le\lambda_2\le\lambda_3$ , that $\lambda_1=1,\;\;\lambda_2=9,\;\;\lambda_3=51$

Now the very last step: from Vieta's formulas we know that $-(-a)=\lambda_1+\lambda_2+\lambda_3$ which leads us to $a=1+9+51=\boxed{61}$

Substitute x=-2 into the polynomial to get (-2)^3 - a(-2)^2 +b(-2) - c = -8 -4a-2b-c = -8 -(4a+2b+c) = -8-1741 = -1749. Let g, h and k be the three roots: (x-g)(x-h)(x-k) (-2-g)(-2-h)(-2-k) = -1749 = (-3)(-11)(-53) This yields g=1, h=9 and k=51 which means p(x) = (x-1)(x-9)(x-51). The value of -a can be deduced by (-1)+(-9)+(-51) = -61 so a=61.

As we can notice, 4a + 2b + c is nothing but a fragment of p(-2) which is equal to -1749. Now, let the roots be x,y,z. -(2+x)(2+y)(2+z) = - 1749 Hence, (2+x),(2+y),(2+z) must be factors of 1749 The factors of 1749 are 3,11,53 Hence, {x,y,z} = {1,9,51} a = x+y+z = 61

let, l,m,n are the roots, so that l+m+n=a, lm+mn+nl=b, lmn=c So, 4a+2b+c=1741, 4(l+m+n)+2(lm+mn+nl)+lmn=1741 (l+2)(m+2)(n+2)=1749=3 11 53 Here, 3, 11 and 53 all are prime numbers. Consider, l+2=3, m+2=11, n+2=53 which gives l=1,m=9,n=51 so that l+m+n=a=61, lm+mn+nl=b=519, lmn=c=459 It is verified that, 4a+2b+c=1741 Therefore, a=61

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 $a = 1 + 9 + 51 = 61$ .

Let the roots be p,q and r The first thing that comes in mind is to find p+q+r which leads to a Now, observe that p(-2)=8-(4a+2b+c)=-1749 Now 1749 can be factorized into 3 * 11 * 53 So p(-2)=-3 11 53=(-2-p)(-2-q)(-2-r) So comparing we get p=1, q=9, r=51 So we want a which is 51+9+1=61

Note that $p(-2) = -8 - 4a - 2b - c = -8 - 1741 = -1749$ . Factoring, $1749 = 3 \times 11 \times 53$ . If $j, k, l$ are the three roots of $p(x)$ , then $p(x) = (x-j)(x-k)(x-l)$ . We can now write $(-2-j)(-2-k)(-2-l) = -1749 = -3 \times -11 \times -53$ . As all the roots are integers, each of $(-2-j), (-2-k),$ and $(-2-l)$ must be an integer.

If any of $(-2-j), (-2-k),$ or $(-2-l)$ are either positive or $-1$ then $j, k,$ or $l$ , respectively, would have to be negative so that subtracting increases $-2$ . This violates the fact that the roots are positive. Therefore $(-2-j), (-2-k), (-2-l)$ must correspond to $-3, -11, -53$ , and $j = 1, k = 9, l = 51$ .

By Viète's root theorem, the sum of an polynomial's roots is the negative of the coefficient of the second term divided by the coefficient of the first term, or in this case $- \frac{(-a)}{1} = a$ . So $a = j + k + l = 1 + 9 + 51 = \boxed{61}$ .

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 $a=x+y+z = 1+9+51=61$

Let the roots of the equation be pqr . So p+q+r=apq+qr+rp=bpqr=c . Since we are dealing with symmetric expressions, WLOG pqr 4a+2b+c=(p+2)(q+2)(r+2)−8 So (p+2)(q+2)(r+2)=1749=31153 Hence (pqr)=(1951) so a=61