You should be able to solve this problem (standard for primary school students in my country where people poo in the...

0 is a positive integer.

nope
0 is non-negative but not positive

m & n are not positive doesn't mean they could only be negative
they could equal 0 too

No. Positive means "greater than zero". Zero is not positive.

easy, add 1 to both sides, so we have m^2 + m + 1 = (n+1)^2
the smallest integer you can add to m^2 to get another square integer is 2m + 1, so m + 1 is too small, therefore, there are no positive integers m and n to solve this equation.

Just set it up as [m(m+1)] / [n(n+2)] = 1 and note that the statement is false for any value of m or n > 0

>then clearly m>n
proof?

>m^2 + m = n^2 + n + n
>hypothetically, if there existed positive integers m and n, prove m>n
need I go on?

>piggots (that's a portmanteau of pig and faggot; it means pig-faggot) ITT giving quasi proofs involving "clear" and "obvious" wild assumptions and random unqualified division

[math]m \neq n[/math] or the first equation doesn't hold. Thus [math]m - n \neq 0[/math].
[eqn]
m(m+1)=n(n+2)\\
m^2+m=n^2+2n\\
m^2-n^2=2n-m\\
(m-n)(m+n)=-2(m-n)\\
m+n=-2\\
[/eqn]

But m and n were positive. Their sum cannot be negative. QED nignogs. Obviously lootards can solve this in primary school, personally I understood this shit in kindergarten but I'm from America the beautiful, so...

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yes, go on, please

suppose, for the sake of contradiction, that m and n are positive integers, and n>m or n = m, and that m^2 + m = n^2 + n + n
there you go, do you want any more?

yes, keep going, all the way to the end

>2n-m = -2(m-n)

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Proof still works.

[math] m^2-n^2=2n-m [/math]

[math] (m-n)(m+n)=-2(m-n)+m [/math]

[math] m+n=-2+\frac{m}{m-n} [/math]

[math] m(1-\frac{1}{m-n})+n=-2 [/math]

Stuff in the paranthesis is greater than zero, or else you can show

[math] m < 1+n [/math]

Contradiction.

It's messier, but it works.

You guys are faggots

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>Tfw had to prove something in partial differential equations and did it graphically but the professor didn't give me credit

Still pissed

How does that indicate what the integer solutions are?

Not a proof.

I feel like induction isn't going to be too laborious to show that no n > m satisfies equation and no m > n satisfies it.

Come on, I'm just pointing out that the problem should say "strictly positive" instead of just "positive". Zero is both positive and negative.

m(m+1) = n(n+2)
can be restated
(m+n)(m-n) = 2n - m.
The RHS has the parity of m because 2n is always even.
Now assume n odd.
Then m+n and m-n both have the opposite parity of m.
This means that (m+n)*(m-n) has the opposite parity of m.
contradiction.
On the other hand assume n even.
Then n(n+2) is even while m*(m+1) is odd. contradiction.
I am too lazy to check for cases where some of the numbers are zero.

Zero is neither positive nor negative

can't be zero. positive integers m and n

soo it's a correct proof?

[eqn]m(m+1)=n(n+2)\\
1m^2+1m+(-2n-n^2)=0\\
m_{1,2}=-1/2\pm(\sqrt{1+4(2n+n^2)})/2\\
n_{1,2}=-1\pm(\sqrt{48})/8 \rightarrow \text{all solutions are non-integers}[/eqn]

i like it

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>m + n = -2
Given that y = m(m+1) and y = n(n+2) are both quadratics that intersect at (m=0,n=0) only, your statement of m + n = -2 cannot be true.

4m^2 + 4m = 4(n^2 + 2n)
(2m+1)^2 = 4(n^2 + 2n) + 1 = 4(n+1)^2 - 3
3 = (2n+2)^2 - (2m+1)^2 = (2n-2m+1)(2n+2m+3)

"proof by picture" only works in Topology

>assume n odd
but n is always even, bcoz m(m+1) is always even

>hurrr the last step of ur proof cant be true!!1
he pointed out the contradiction brainlet

>didn't show that you understood the material required for the exam
>wtf why'd i get marked wrong for this

Assume that m^2 + m = n^2 + 2n holds true for some positive integers. Assume for the sake of contradiction that n>m.

We know that n^2 + n < n^2 + 2n so m^2 + m > n^2 + n rearranging we get.

m^2 - n^2 > n-m

This itself is a contradiction since |n| > |m|,then m^2 - n^2 should be negative but n - m is positive.

We can be more verbose though, since n-m = -(m-n) and m-n is negative so diving by it will flip the sign, we get.

(m+n)(m-n) > -(m-n)

So m+n < -1

Clearly a contradiction.

We can also look at it like this.

n>m

n^2 > m^2 given that n,m are positive.

n-m > 0 > m^2 - n^2

But that's impossible because we know that m^2 - n^2 > n-m

>Suppose equality holds
Prove it directly, fag.

You can rearrange it to get m/n = n+2/m+1.
We know m cannot equal n since that wouldn't make sense.
So it is either larger or smaller.
If m>n then n+2>m+1
Rearranging we get 1>m-n which can not be possible if m and n are natural numbers.
If n>m then m+1>n+2
Rearranging we get m-n>1 which is not possible since n>m.

You can rearrange it to get m/n = n+2/m+1.
We know m cannot equal n since that wouldn't make sense.
So it is either larger or smaller.
If m>n then n+2>m+1
Rearranging we get 1>m-n which can not be possible if m and n are positive integers.
If n>m then m+1>n+2
Rearranging we get m-n>1 which is not possible since n>m.
This all tells us there is no solution involving positive integers.

and for a good reason too
---
hypotenuse length, [math] n \rightarrow \infty [/math]

a+b or [math] \sqrt{a^{2}+b^{2}} [/math] ?

Attached: hypotenuse length.png (203x191, 2K)

Add 1 to both sides and complete the square
[math](m+\frac12)^2+\frac34 = (n+1)^2[/math].
LHS must be odd, so
n must be even
which means you can factor a 2 from the RHS of the starting equation giving
[math]m(m+1)=4\dot n ( \dot n +1)[/math].
implying [math]m =4 \dot n =\dot n.[/math] a contradiction for positive numbers

>the value of m can be solved with the quadratic formula as a function of n
>the solutions of 0 = 1 + 4(2n + n^2) are irrational, therefore all pairs of numbers solving m(m+1) = n(n+2) are not pairs of integers.
what did he mean by this?

Best proof in thread

Very nice.
m^2 < m^2+m+1 < (m+1)^2