2
$\begingroup$

After simply factorizing the equation given in the question I got $\mathbf{m + n = -0.5}$ . But the question mentioned $m, n ∈ \mathbb{N}$ . Then how $m + n = -0.5$ ?

Did I do some mistake or is the question wrong?

$\endgroup$
7
  • 2
    $\begingroup$ There's the second factor. What happens if that is $0$? $\endgroup$ Nov 13 '20 at 14:07
  • 1
    $\begingroup$ What do u mean by "There's the second factor" $\endgroup$ Nov 13 '20 at 14:12
  • 2
    $\begingroup$ You wrote "factorizing". I suppose you got more than one factor doing that. $\endgroup$ Nov 13 '20 at 14:13
  • $\begingroup$ Oh thanks that second factor gives m = n and solves the question ;) $\endgroup$ Nov 13 '20 at 14:17
  • $\begingroup$ How can i end my question ? $\endgroup$ Nov 13 '20 at 14:18
5
$\begingroup$

Yes. $m-n$ is a perfect square.

Just observe that your condition implies that $$ (m-n)(2m+2n+1)=0$$ Since $m$ and $n$ are natural numbers, we must have $m-n = 0$ which is a perfect square

$\endgroup$
3
  • $\begingroup$ Not clear precisely which way you deduced $\,m=n,\,$ but if it is from $\,m\neq n\Rightarrow\, -2(m+n) = 1\,$ then this proof by parity (divisibility) contradiction generalizes widely - see my answer where I shows how to extend it from the divisor $2$ to $d.\ \ $ $\endgroup$ Nov 14 '20 at 2:14
  • $\begingroup$ @Bill Dubuque product of two integers is 0 implies that atleast one of them is 0. Since $m$ and $n$ are naturals, $2m+2n+1$ cannot be $0$. $\endgroup$ Dec 12 '20 at 13:50
  • $\begingroup$ Yes, that's essentially the proof by parity that I surmised you used. $\endgroup$ Dec 12 '20 at 15:17
4
$\begingroup$

As you conclude in the comments, the point is that we must have $m = n$.

Here's an alternative proof: suppose for contradiction that $m \neq n$; in particular take $m > n$. Let $k = 2m^2 + m$. We see that $m$ and $n$ are two solutions to the quadratic equation $$ 2x^2 + x - k = 0. $$ The sum of the roots of $ax^2 + bx + c = 0$ is given by $-\frac{b}{a} = - \frac 12$, which is to say that $m + n$ is not an integer. This is impossible since $m,n \in \Bbb Z$.

$\endgroup$
6
  • $\begingroup$ Nice method but u just complicated it ,if x is not equal to y then x + y = -0.5 which is not true $\endgroup$ Nov 13 '20 at 14:26
  • $\begingroup$ @AdhirajSinghBrar You're right. When I first wrote it I was trying to figure out why $m-n$ would be a perfect square... $\endgroup$ Nov 13 '20 at 14:27
  • $\begingroup$ @AdhirajSinghBrar I made the proof more succinct now $\endgroup$ Nov 13 '20 at 14:31
  • $\begingroup$ @Adhiraj I added an answers which emphasizes the underlying number theory (per your tag). $\endgroup$ Nov 14 '20 at 2:11
  • $\begingroup$ Thanks @BenGrossmann $\endgroup$ Nov 14 '20 at 3:57
2
$\begingroup$

An alternative way to approach the same result: For nonnegative values of $x$, the function $f(x)=2x^2+x$ is strictly increasing, so the only way to have $f(m)=f(n)$ is if $m=n$. And then of course $m-n=0$ is a perfect square.

$\endgroup$
2
$\begingroup$

Special case of that below, namely $\, d=2,\ f = 2x^2\!+\!x$.

Theorem $\ $ If $\,f(x)\in\Bbb Z[x]\,$ and $\!\bmod d\!:\ f\equiv bx\!+\!c,\, \color{#c00}{b\not\equiv 0}\,$ then $\,f(m)=f(n)\Rightarrow m = n$

Proof $\,\ 0 = f(m)\!-\!f(n) = (m\!-\!n)\:\!k,\,$ for $\,k = \dfrac{f(m)-f(n)}{m-n}\in\Bbb Z,\,$ thus if $\,m\neq n\,$ then $\,k = 0\,$ thus $\bmod d\!:\,\ \color{#c00}0 = k \equiv \dfrac{bm\!+\!c-(bn\!+\!c)}{m-n}\equiv \color{#c00}b,\,$ contra hypothesis.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.