# Homework Help: How to disprove (m^2)+m+1=(n^2)?

1. Feb 26, 2017

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Prove: There exists no $m,n∈ℕ$ such that $m^2+m+1=n^2$."

2. Relevant equations

3. The attempt at a solution
I basically rewrote it as:

$m^2+2m+1=n^2+m$

or

$(m+1)^2=n^2+m$,

and subtracting $n^2$, I get

$(m+1-n)(m+1+n)=m$.

Then I divided both sides of the equation by $m$ to get:

$(1+\frac{1}{m}-\frac{n}{m})(m+n+1)=1$.

I then argued that the only way for this to be possible was (1) if both terms on the left were $1$, which would contradict the fact that $m$ and $n$ were positive integers, because if they were, the second term would be greater than one; or (2) if the two terms were multiplicative inverses of each other. The latter is the one I had trouble with, because it led to some kind of circular reasoning. Basically, I ended up with the exact same problem which I had to once again disprove. Can anyone provide me with any hints on how to approach this proof?

2. Feb 26, 2017

### Staff: Mentor

I have no idea how to proceed from the point you got stuck. But I think, it could be done by solving $m^2+m+(1-n^2)=0$ for $m$ and then analyze the root. Also consider, that beside the root, the summand $-\frac{1}{2}$ that you get has to vanish, too.

3. Feb 27, 2017

### pasmith

Set $m^2 + m + 1 - n^2 = (m - m_1)(m - m_2)$. What must $m_1 + m_2$ be?

4. Feb 27, 2017

### PeroK

Here's another idea. It seems to me that adding $m +1$ is not enough to get you from $m^2$ to the next square.

5. Feb 27, 2017

### Eclair_de_XII

Thanks for the help. I finished it, by the way.