0

I have got two integers, say $m$ and $n$, which I’m trying to guess. As a hint I’m getting the value of $mn+11m+11n-m^2-n^2$. After that I can unambiguously name $m$ and $n$. What was the value of the expression? In other words I need to find such unique pair of $(m,n)$, for which this value is also unique. My attempt was to transform expression $mn+11m+11n-m^2-n^2 = a$, where $a \in \mathbb{Z}$ and $m, n \in \mathbb{R}$ to canonical form so we can get one of the 5 curves of 2nd order I.e. ellipse, point, empty set, pair of intersected lines and hyperbola. Empty set is not our case obvious. Ellipse, hyperbola and lines are not either simply because we can get one more pair out of one (just one of them negative for example). For our case we only can have ellipse, point, or empty set as one of invariant is positive. So we need to calculate the third one to get the answer

Moni145
  • 2,142
BeesaFangirl DOTO
  • 436
  • 2
  • 8
  • What contest is this from? – saulspatz Aug 19 '21 at 00:47
  • I don’t think that this problem came from actual contest. My friend just gave me this problem. But I’m not sure about the category of this task, so I put contest-math for it, because I thought it fit nice for this type of problems. – BeesaFangirl DOTO Aug 19 '21 at 00:56
  • Since the given polynomial is symmetric in $m$ and $n$, there can't be a unique answer unless it turns out that $m=n$. Then the polynomial becomes $m(22-m)$, so for a unique answer we need $n=m=11$ and the mystery value can only be $121$. Now you have to show that $121$ actually works. – saulspatz Aug 19 '21 at 01:19
  • I think the problem means uniqueness up to permutation of $m$ and $n$ – BeesaFangirl DOTO Aug 19 '21 at 01:22
  • Also I come up with idea to transform this expression to canonical form so I can get some surface of second order – BeesaFangirl DOTO Aug 19 '21 at 01:27
  • 1
    Can you include your efforts for this problem per MSE policy? – Calvin Lin Aug 19 '21 at 01:28

2 Answers2

2

Let $f(m,n)$ be the expression.

Hints:

  • For a fixed value of $m$, viewing the expression as a quadratic in $n$,
    $n^2 -(m+11)n + m^2-11m = k $ has 2 roots.
    If $n^*$ is one of the roots, then what is the other root in terms of $n^*$ and $m$?
  • Hence $f(m, n) = f(m , n')$ for what value of $n'$ in terms of $n$ (and $m$)?
  • Hence, the expression is NOT unique if $ n \neq n'$.
  • Repeat for $m$.
  • Conclude that the expression is NOT unique unless $ n = m = 11$.
  • Verify that $f(m, n) = 121$ has only 1 solution.
Calvin Lin
  • 68,864
1

If $a=m-11$ and $b=n-11$, the expression is equivalent to $$ab-b^2-a^2+121$$ Hence, we need to find a $k\in\mathbb{Z}$ s.t. $$a^2+b^2-ab=k$$ has a unique integer solution.

It is clear that for any integer solution $(a_0,b_0)$, we also have that $(-a_0,-b_0)$ is another integer solution. Hence there only exists a unique solution if $$\begin{cases}a_0=-a_0\\b_0=-b_0\end{cases}\implies\begin{cases}a_0=0\\b_0=0\end{cases}$$

When $a,b=0$ we have $k=0$. To verify that $a^2+b^2-ab=0$ has only $1$ real solution, note that if $ab\geq 0$, then $$a^2+b^2-ab$$ $$=(a-b)^2+ab\geq 0$$ This has equality only when $ab=0$ and $a=b$, which only occurs when $a=b=0$

If $ab\leq 0$, then $$a^2+b^2-ab\geq 0$$ This has equality only when $a=b=ab=0$.

Hence, the unique solution is when $a,b=0$. Our expression evaluates to $\boxed{121}$

Alan Abraham
  • 5,142
  • 6
  • 20
  • You mean to say that $(-a_0,-b_0)$ is another solution, not $(-a_0,b_0)$ – saulspatz Aug 19 '21 at 09:07
  • Oops, that was a typo. – Alan Abraham Aug 19 '21 at 12:22
  • 1
    Note that what you have is "If $F(a_0, b_0) = k $, then $F(-a_0, -b_0) = k$." which establishes that a certain solution is not unique. You do not yet have "If $F(a_0, b_0) = k $, then $F(a^, b^ ) = k $ iff $ a^* = \pm a, b^* = \pm b$" which is what you need to establish that a solution is indeed unique. This is why I have a final step of "Verify that $f(m, n) = 121$ has only 1 solution." – Calvin Lin Aug 19 '21 at 16:35
  • That is true, I'll edit my solution – Alan Abraham Aug 19 '21 at 17:05