An Interesting Expansion Product

Number Theory Level 5

( x + 1 ) p ( x − 3 ) q = x n + a 1 x n − 1 + a 2 x n − 2 + ⋯ + a n \begin{aligned} (x+1)^p(x-3)^q = x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_n \end{aligned} Consider the expansion above where a 1 , a 2 , ⋯ , a n , p a_1, a_2, \cdots, a_n, p and q q are integers with p p and q q positive and n = p + q n = p+q . How many ordered pairs ( p , q ) (p, q) with 1 ≤ p , q < 1000 1\leq p, q < 1000 are there such that a 1 = a 2 a_1 = a_2 ?


The answer is 19.

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Russelle Guadalupe by Russelle Guadalupe , 25, Philippines

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1 solution

Chew-Seong Cheong
Jun 16, 2015

By Vieta's formulas, we have:

a 1 = − [ ( − 1 ) p + ( 3 ) q ] = p − 3 q a 2 = ( p 2 ) ( − 1 ) ( − 1 ) + p q ( − 1 ) ( 3 ) + ( p 2 ) ( 3 ) ( 3 ) = p ( p − 1 ) 2 − 3 p q + 9 q ( q − 1 ) 2 a 1 = a 2 ⇒ p − 3 q = p ( p − 1 ) 2 − 3 p q + 9 q ( q − 1 ) 2 2 p − 6 q = p 2 − p − 6 p q + 9 p 2 − 9 q 0 = p 2 − 6 p q + 9 p 2 − 3 p − 3 q ⇒ ( p − 3 q ) 2 = 3 ( p + q ) \begin{aligned} a_1 & = -[(-1)p+(3)q] = p-3q \\ a_2 & = \begin{pmatrix} p \\ 2 \end{pmatrix} (-1)(-1) + pq (-1)(3) + \begin{pmatrix} p \\ 2 \end{pmatrix} (3)(3) \\ & = \dfrac{p(p-1)}{2} -3 pq + \dfrac{9q(q-1)}{2} \\ a_1 & = a_2 \\ \Rightarrow p - 3q & = \frac{p(p-1)}{2} -3 pq + \frac{9q(q-1)}{2} \\ 2p-6q & = p^2 - p - 6pq + 9p^2 - 9q \\ 0 & = p^2 - 6pq + 9p^2 - 3p - 3q \\ \Rightarrow (p-3q)^2 & = 3(p+q) \end{aligned}

For integer p p and q q , ( p − 3 q ) 2 = 3 ( 3 m 2 ) (p-3q)^2 = 3(3m^2) , where m m is a positive integer.

⇒ { p − 3 q = ± 3 m . . . ( 1 ) p + q = 3 m 2 . . . ( 2 ) ⇒ { ( 2 ) − ( 1 ) : q = 3 m ( m ± 1 ) 4 ( 2 ) : p = 3 m ( 3 m ∓ 1 ) 4 \Rightarrow \begin{cases} p-3q = \pm 3m &...(1) \\ p+q = 3m^2 &...(2) \end{cases} \\ \Rightarrow \begin{cases} (2)-(1): & q = \dfrac {3m(m\pm 1)}{4} \\ (2): & p = \dfrac{3m(3m \mp 1)}{4} \end{cases}

The ordered pairs of ( p , q ) (p,q) are as follows:

m p = 3 m ( 3 m − 1 ) 4 q = 3 m ( m + 1 ) 4 p = 3 m ( 3 m + 1 ) 4 q = 3 m ( m − 1 ) 4 3 18 9 4 33 15 39 9 5 60 15 7 105 42 8 138 54 150 42 9 189 54 11 264 99 12 315 117 333 99 13 390 117 15 495 180 16 564 204 588 180 17 663 204 19 798 285 20 885 315 915 285 \begin{array} {ccccc} m & p = \frac{3m(3m-1)}{4} & q = \frac{3m(m+1)}{4} & p = \frac{3m(3m+1)}{4} & q = \frac{3m(m-1)}{4} \\ 3 & 18 & 9 & & \\ 4 & 33 &15 & 39 & 9 \\ 5 & & & 60 & 15 \\ 7 & 105 & 42 & & \\ 8 & 138 & 54 & 150 & 42 \\ 9 & & & 189 & 54 \\ 11 & 264 & 99 & & \\ 12 & 315 & 117 & 333 & 99 \\ 13 & & & 390 & 117 \\ 15 & 495 & 180 & & \\ 16 & 564 & 204 & 588 & 180 \\ 17 & & & 663 & 204 \\ 19 & 798 & 285 & & \\ 20 & 885 & 315 & 915 & 285 \end{array}

There are 19 \boxed{19} ordered pairs of ( p , q ) (p,q) such that a 1 = a 2 a_1=a_2 .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Great job.

For a complete characterization of solutions to ( p − 3 q ) 2 = 3 ( p + q ) (p-3q)^2 =3 (p+q) , you should also explain which values of m m will lead to an integer solution. The list that you create suggests that the answer is dependent on m ( m o d 3 ) m \pmod{3} .

Nice job, I kept missing the largest case :(. The following made finding all the cases a lot easier. If you complete the square such that the linear term only contains one variable, the cases can be combined very nicely: ( p − 3 q ) 2 = 3 ( p + q ) = 3 ( p − 3 q ) + 12 q ⇒ X 2 : = ( 2 p − 6 q − 3 ) 2 = 48 q + 9 ∣ ⋅ 4 , + 9 \begin{aligned} (p-3q)^2 &= 3(p+q) = \red{3}(p-3q) + 12q & \Rightarrow && X^2:=(2p-6q - \red{3})^2 &= 48q+9&&&\left|\cdot 4,\quad + 9\right. \end{aligned} We notice X 2 X^2 is non-zero and dividsible by 3 3 , so we can set X = 3 m > 0 X=3m>0 : ± 3 m = 2 p − 6 q − 3 9 m 2 = 48 q + 9 } q = 3 ( m 2 − 1 ) 16 , p 1 , 2 = 3 2 ( 2 q + 1 ∓ m ) , m > 0 \begin{aligned} \left.\begin{aligned} \pm 3m &= 2p-6q - 3\\ 9m^2 &= 48q + 9 \end{aligned}\right\} && q &= \frac{3(m^2-1)}{16}, &&& p_{1,2} = \frac{3}{2}(2q + 1 \mp m),\quad m>0 \end{aligned} Checking quadratic residues m o d 16 \mod 16 , we only get integer solutions for q q if we choose m = 8 k ± 1 m=8k\pm 1 . Let's list them until p , q > 1000 p,\:q>1000 : m = 8 k − 1 p 1 p 2 q m = 8 k + 1 p 1 p 2 q 7 18 39 9 9 33 60 15 15 105 150 42 17 138 189 54 23 264 333 99 25 315 390 117 31 495 588 180 33 564 663 204 39 798 915 285 41 885 1008 315 47 1173 1314 414 \begin{array}{r|rrr||r|rrr} m=8k- 1 & p_1 & p_2 & q & m=8k+1 & p_1& p_2 &q\\\hline 7 & 18 &39 &9 & 9 & 33&60&15\\ 15 & 105&150&42 & 17 & 138&189&54\\ 23 & 264&333&99 & 25 & 315&390&117\\ 31 & 495&588&180 & 33 & 564&663&204\\ 39 & 798&915&285 & 41 & 885&\red{1008}&315\\ 47 & \red{1173} &\red{1314} & 414 & & \end{array} Excluding the pairs with p > 1000 \red{p>1000} , we have 19 \boxed{19} distinct solutions! 

/* maxima code to find all solutions */
result(m) := [
    3 / 16 * (3*m^2 - 8*m + 5), /* p1 */
    3 / 16 * (3*m^2 + 8*m + 5), /* p2 */
    3 / 16 * (m^2-1)            /* q */
]$

for m : 2 thru 47 do (
    if mod(m^2 - 1, 16) = 0 then disp(m, result(m))
)$

Carsten Meyer - 1 month, 4 weeks ago

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