Samir's Cubic Roots

$r+s+t=a$ (By the Vieta's formula )

$(r+s)(s+t)(r+t)=(a-r)(a-s)(a-t)$

$P(x)$ has roots $r, s,$ and $t$

$\implies P(a-x)$ has roots $a-r, a-s$ and $a-t$

Let $P(a-x)=c_3 x^3 + c_2 x^2 + c_1 x^1 + c_0$ . By the Vieta's formula , $(a-r)(a-s)(a-t)=- c_3 / c_0$ , so we only need to calculate $c_3$ and $c_0$ .

Because the coefficient of $x^3$ in $P(x)$ is $1, c_3 = -1$

$c_0 = P(a-0) = P(a) = - a^3 b^2 - 9 a^2 b^2$

Therefore $(r+s)(s+t)(r+t)=(a-r)(a-s)(a-t)=- c_3 / c_0 = a^3 b^2 + 9 a^2 b^2 = a^2 b^2 ( ab + 9 ) = k^2$

$a$ and $b$ are positive integers, so $ab+9$ is also a square of a positive integer. Let $ab+9=l^2$ . Then $ab=(l+3)(l-3)$ . we can easily find that $(a, b)=(31, 25)$ can be a solution of the equation. In this case, $l=28$ .

If $l>28$ , we can show that there is no solution of $ab+9=l^2$ in the range(If $l=29, 30$ or $31$ , use direct computation. If $l \geq 32, ab>31 \cdot 31$ so there is no solution).

Therefore, The possible maximum value of $ab$ is $31\times 25 = 775$ .

Using Vièt theorem, we get: $\left\{\begin{array}{l}r+s+t=a\\rs+st+rt=a^2b^3\\rst=-9a^2b^2\end{array}\right.$ .

Hence, $k^2=(r+s)(s+t)(r+t)=(r+s+t)(rs+st+rt)-rst$ $=a^3b^3+9a^2b^2=a^2b^2(ab+9).$

This implies to $ab+9$ is a perfect square, put: $ab+9=n^2$ .

Since $1\le a\le 31;1\le b\le 31$ , $n^2=ab+9\le 970 \Rightarrow n\le31$ .

If $n=31$ , $ab=n^2-9=952$ , there are not exist numbers $a,b$ such that.

If $n=30$ , $ab=n^2-9=891$ , there are not exist numbers $a,b$ such that.

If $n=29$ , $ab=n^2-9=832$ , there are not exist numbers $a,b$ such that.

If $n=28$ , $ab=n^2-9=775$ , we can choose: $a=25,b=31$ .

So, the maximum possible value of $ab$ is $775$ .

We know that $r+s+t=a$ , so $(r+s)(s+t)(r+t)=(a-r)(a-s)(a-t)=P(a)$ since $P(a)$ is monic, $P(a)=a^3-a^3+a^3b^3+9a^2b^2=a^2b^2(ab+9)=k^2$ which means that $ab+9$ is a perfect square, $ab=c^2-9$ for positive integer $c$ . Since $a,b$ have a upper bound of 31, we try $c=31, 30, 29 \ldots$ , since $32^2-9>31^2$ . Without loss of generality, let $a \geq b$ . If we choose the divisor $d \geq \sqrt{c^2-9}$ of $c^2-9$ that is closest to $\sqrt{c^2-9}$ . If $d>31$ , obviously there are no solutions since $31\geq a \geq \sqrt{c^2-9}$ . For $c=31, d=34$ . For $c=30, d=33$ . For $c=29, d=32$ . For $c=28, d=31$ . Therefore $a=31, b=25, ab=775$ .

Change (r+s)(s+t)(t+r)=k^2 as (a-r)(a-t)(a-s)=k^2 From polinomial we get: r+s+t=a ...(1) rs+st+rs=a^2b^2 ...(2) rst=9a^2b^3 ...(3) from 1,2,3 and also we get (a-r)(a-t)(a-s)=(ab)^2(ab+9)=k^2 ab+9 must be quadratic number... Now suppose that ab+9=l^2 now we strart from l=31 because 0<=a,b<=31. for l=31, there is not a,b for l=30, there is not a,b for l=29, there is not a,b and for l=28, there is a,b=25,31. If l maximum value then ab maximum value... and we get 775

$(r+s)(s+t)(r+t) = (a-r)(a-s)(a-t) = P(a)$

$P(a)=a^3-a^3+a^3b^3+9a^2b^2 = a^2b^2(9+ab) = k^2$

thus $9+ab$ is a perfect square

$9 + ab = n^2$

$ab = (n+3)(n-3)$

$n=28$ works, $n=29~31$ doesn't.

We have:

r+s+t = a

rs + sr + rt = (a^2)(b^3)

rst = -9(a^2)(b^2)

(r+s)(s+t)(r+t) = (rs + st + rt)(r+s+t) - rst

k^2 = (a^3)(b^3) + 9a^2b^2

Let k = (ab)(m)

m^2 = ab + 9

(m-3)(m+3) = ab

Observe that m < 32 since 32^2 - 3 > 961

m = 31: (28)(34) = ab

• Solutions for a and b cannot be found as 34 is a multiple of 17; the only multiple of 17 in (1,31) is 17, and (28)(34)/17 = 56 > 31

m = 30: (27)(33) = ab

• Again no solutions; the only primes in (27)(33) are 3 and 11; if a is a multiple of 11 it cant also be a multiple of 3, and (27)(33)/11 = 81 > 31

m = 29: (26)(32) = ab

• No solutions; only prime factors are 2 and 13; if a is a multiple of 13 it cannot be a multiple of 4; (26)(32)/26 = 32 > 31

m = 28: a = 31, b = 25 is a solution, giving us ab = 775

interestingly, it is easily provable that r,s,t cannot be simultaneously real:

r^2 + s^2 + t^2 + 2rs + 2st + 2rt = a^2

r^2 + s^2 + t^2 - rs - st - rt = 1/2 { (r-s)^2 + (r-t)^2 + (s-t)^2 }, implying

r^2 + s^2 + t^2 >= rs + st + rt

a^2 = r^2 + s^2 + t^2 + 2rs + 2st + 2rt >= 3rs + 3 st + 3rt = 3a^2b^3

a^2 >= 3a^2b^3

1 >= 3b^3; since b is a positive integer, contradiction.

Because $P(x)=x^3 - a x^2 + a^2 b^3 x + 9 a^2 b^2$ , then $r+s+t=a$ .

Hence, $(r+s)(s+t)(r+t) = (a-t)(a-r)(a-s)$ .

Meanwhile, $P(x)$ can be wrote as $P(x)= (x-t)(x-r)(x-s)$ .

Substituting $x$ with $a$ , now $P(a)= (a-t)(a-r)(a-s)$ .

Because $(r+s)(s+t)(r+t) = k^2$ , then $P(a)=k^2$ , which means $a^3 - a\cdot a^2 + a^2 b^3\cdot a + 9 a^2 b^2= k^2$ .

Thus we get $a^2 b^2 (ab + 9)= k^2$ .

$a$ , $b$ , $k$ are positive integers, so $ab+9$ must equal some $m^2$ .

$ab\leq 31^2$ , and $ab=(m+3)(m-3)$ , just need to verify $m$ when $m+3\leq 31$ .

At last, $ab=(31)(31-3-3)=31\times 25=32\times 25 - 25=800 - 25 =775$ .

First. From Vieta theory we can get :

1) $r + s + t = a \Rightarrow$ (1)

2) $rs + rt + st = a^2b^3 \Rightarrow$ (2)

3) $rst = -9a^2b^3 \Rightarrow$ (3)

$(r+s)(s+t)(t+r) = (r+s)(rs) + (r+t)(rt) + (s+t)(st) + 2rst \Rightarrow$ (4)

from (1) we can get :

$r + s = a - t,$

$r + t = a - s,$

$s + t = a - r.$

so, (4) can be changed into :

$(a-r)(st) + (a-t)(rs) + (a-s)(rt) + 2rst.$

$(a)(rs + st + rt) -3rst + 2rst.$

$(a)(rs + st + rt) - rst. \Rightarrow$ (5)

from (2) and (3), we can change (5) into :

$(a)(a^2b^3) - (-9a^2b^2).$

$((ab)^2)(ab + 9).$

we know that $(r+s)(s+t)(r+t) = k^2.$ with $k$ is a positive integer.

$(ab+9) = (\frac{k}{ab})^2.$

We can assume $(\frac{k}{ab})^2 = w^2.$

So, we get $(ab+9) = w^2$

$ab = w^2 - 3^2$

$ab = (w-3)(w+3)$

Remember that $a_{max} = 31$ and $b_{max} = 31.$

$ab$ become max if $w+3 = a_{max}$ or $b_{max}.$

$ab_{max} = 31*(31-6) = 775.$

Using Vieta we can conclude that

$r+s+t = a$

$rs+st+tr = a^2b^3$

$rst = 9a^2b^2$

then $(r+s)(s+t)(t+r) = (a-t)(a-r)(a-s) =k^2$

so $a^3-(r+s+t)a^2+(rs+st+tr)a+rst$

$= a^3 - a^3 + a^3b^3 + 9a^2b^2 = k^2$

for k $\in \mathbb{N}$

and $a^2b^2 (ab+9)=k^2$ so we find the maximum value of ab

where $ab+9 = l^2$ where $l \in \mathbb{N}$

$ab+9 =l^2$

$ab = l^2-9 = (l-3)(l+3)$

then the maximum value will be reached when $ab=25 \times 31 = 775$

since r, s and t are roots of $x^3-ax^2+a^2b^3x+9a^2b^2$ , we have $r+s+t = a;$ $rs+rt+st = a^2b^3;$ $rst = -9a^2b^2;$ hence $(r+s)(s+t)(r+t)$ $= (rs+rt+st)(r+s+t) - rst$ $= a^3b^3+9a^2b^2$ $= a^2b^2(ab+9);$ On the other hand, since $a^2b^2(ab+9)$ is a square of a integer, and $ab+9, a, b$ are integers, then it can be easily seen that $ab+9$ is also a perfect square, which means that $ab = u^2 - 9$ for some integer $u \leq 31$ . By a few trials, we find that $u = 25$ is the largest integer that satisfies that $u^2-9$ can be factored as the product of two integers $\leq 31$ . Hence the maximum possible value of $ab$ is $25^2-9 = 775$ ;

hee, hee.

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from numpy import real, imag, roots
from sympy.ntheory.primetest import is_square

ab_max = -10**6
for a in range(1, 32):
    for b in range(1, 32):
        coeff = [1, -a, a**2 * b**3, 9*a**2 * b**2]
        r, s, t = roots(coeff)
        prod = (r + s) * (s + t) * (r + t)
        if abs(imag(prod)) < 10**-8 and is_square(int(round(real(prod), 8))):
            ab = a*b
            if ab > ab_max:
                ab_max = ab
print(ab_max)

1

775