$m$ and $n$ all the way !!!

Number Theory Level 5

The number of ordered pairs of integers $(m,n)$ such that $mn ≥ 0$ and $m^3+n^3+99mn=33^3$ is

The answer is 35.

4 solutions

Michael Tong
Dec 15, 2013

I see two variables, third powers, and coefficients related to $3$ , so I immediately think of

$(m + n)^3 = m^3 + 3m^2n + 3mn^2 + n^3$

So, we need terms of $3m^2n$ and $3mn^2$ . Well, if we let $m + n = 33$ , then we get

$m^3 + 3(m + n)(mn) + n^3 = (m + n)^3$ , which fits our identity. So we get a family of $34$ integer solutions in $m + n = 33$ .

Well, so now we know that $m + n - 33$ divides $m^3 + n^3 + 99mn - 33^3$ evenly. Doing multi-variable polynomial long division, we get a neat factorization of $m^3 + n^3 + 99mn - 33^3 = (m + n - 33)(m^2 + n^2 - mn + 33m + 33n + 33^2)$ . Now, I see squares, the term $mn$ , and related coefficients. So I think of $(a+b)^2 = a^2 + 2ab + b^2$ and factorize the second term into $\frac{1}{2} ((m - n)^2 + (m + 33)^2 + (n + 33)^2)$ . This is set equal to $0$ , and since all terms are squared then they all must equal zero. This gives a solution of $(-33, -33)$ , giving us a total of $\boxed{35}$ solutions, and we are done.

Well i approached the next part of getting a $(-33 , -33)$ in a different way. Hows this? If $m,n \geq 0$ we have 34 solutions. Now suppose $m=-k , n=-l$ So u have, $k^3 + l^3 + 99kl = 33^3$ and

$-k^3 - l^3 -3kl(k+l) = -(k+l)^3$ .

Now subtract the second from the first and assuming $k+l \neq -33$ and after some simplification we get

$k^2 + l^2 - kl - 33k - 33l + 33^2 = 0$

$\implies$ $k^2 -(l+33)k + (l^2 - 33l + 33^2) = 0$ .

So $k = \dfrac{l+33 \pm \sqrt{(l+33)^2 - 4(l^2 - 33l+33^2)}}{2}$

k is positive and so we have

$(l+33)^2 \geq 4(l^2 - 33l+33^2)$ $\implies$ $66l \geq l^2 + 33^2$ .

As $l$ and $33$ are positive, we have $l^2 + 33^2 \geq 66l$ . So it implies the inequality turns into an equality and we have $l= - 33$ and plugging it into the equation gives $k = 33$ and so the only solution when $m,n \leq 0$ is $(m,n) = (-33,-33)$

Sagnik Saha - 7 years, 3 months ago

Perfect!! Very well explained.

Dinesh Chavan - 7 years, 5 months ago

Dinesh Chavan, this is Problem 30 from ASHME 1999. Please don't blatantly copy problems from another competition.

Vinayak Kumar - 7 years, 5 months ago

I liked the format of this question. Also many people don's know about this exams.So to make the people aware of such questions I posted this.I meant no genuine possession of this problem.Many people solved it like a mere challenge.Yet I will take your advice and take care of it next time.Thanks for your advice

Dinesh Chavan - 7 years, 5 months ago

@Dinesh Chavan – Yeah,that thing needs to be mentioned in the problem always, you must say it is taken from which contest... Agreed!!!

Aditya Raut - 6 years, 11 months ago

I encountered this problem in the book 101 algebra problems...

Xuming Liang - 7 years, 5 months ago

Nicely done.

Nishant Sharma - 6 years, 11 months ago

Andrew Ong
Dec 15, 2013

$m^3+n^3+99mn = 33^3 \\ \Rightarrow m^3+n^3-33^3+99mn = 0 \\ \Rightarrow (m+n-33)(m^2+n^2+33^2-mn+33m+33n) = 0 \\ \Rightarrow \frac{1}{2}(m+n-33) [(m-n)^2+(m-33)^2+(n-33)^2] = 0 \\ \Rightarrow m+n-33=0 \cup m=n=33$ This gives us 34 and 1 possibilities, respectively, for a total of $\boxed{35}$

$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)$

Andrew Ong - 7 years, 5 months ago

The step before last would be (m-n)^2+(m+33)^2+(n+33)^2=0 so we get (-33,-33) instead of (33,33)

Kuladip Maity - 7 years, 5 months ago

there is an easier way...

(x+y)^{3}=x^{3}+3x^{2}y+3y^{2}x+y^{3}

x^{3}+3xy(x+y)+y^{3}

so for 3xy(x+y)=99, (x+y) needs to be 33 ,

and (x+y)^{3} will be 33^{3}

Whatever 21 - 7 years, 5 months ago

How did you possibly know to factor it like that?

Mike Kong - 7 years, 5 months ago

Deepak Kamlesh
Dec 18, 2013

We are given that

m^3 + n^3 + 99mn = 33^3

This implies that

m^3 + n^3 + (-33)^3 = 3(-33)mn,

i.e. a^3 + b^3 + c^3 = 3abc

where a=m, b=n, c= -33.

Now, a^3 + b^3 + c^3 - 3abc = 1/2 (a+b+c) [(a-b)^2 + (b-c)^2 + (c-a)^2]

So, a^3 + b^3 + c^3 = 3abc if and only if either a+b+c=0 (i.e. m + n = 33)

or a = b = c (i.e. m = n = -33)

Now, m + n = 33 has 34 solutions which are (0,33), (1,32),...(32,1),(33,0).

Also, m = n = -33 is a solution.

So, in all we have 35 solutions i.e. 35 ordered pairs with the required property.

Jinay Patel
Dec 17, 2013

Note that (m + n)^3 = m^3 + n^3 + 3 m n(m+n) If m + n = 33 , then 33^2 = (m+n)^3 = m^3 +n^3 + 3 m n (m+n) = m^3 + n^3 + 99 m n Hence , m+n-33 is a factor of m^3 + n^3 + 99 m n -33^2. we have m^3 + n^3 + 99 m n -33^2 = (m+n-33)(m^2 + n^2 -m n +33 m + 33 n + 33^2) = (m+n-33)[(m-n)^2 + (m+33)^2 + (n+33)^2]/2 Hence there are 35 solutions altogether.

m m m and n n n all the way !!!

The answer is 35.

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$m$ and $n$ all the way !!!