$(m^2+n^2) \div (m+n)$

Here's a visual proof.

Let $m^2$ and $n^2$ represent two squares. Their combines area is $m^2 + n^2$ as shown below.

The total length of the bottom is $m + n$ . If we multiply that value by m, we get the following blue area.

However, that's not the total area of $m^2 + n^2$ , because it misses the $1 * n$ rectangle in the top right. So maybe we just need to multiply $m + n$ by one more. Since m and n are consecutive positive integers, the next integer is n. When multiplying $m + n$ by n, we get the following area:

The area of $(m + n) * n$ is the blue area and the red area combined. This also isn't the total area of $m^2 + n^2$ , because it overshoots the area by the red rectangle of $1 * m$ .

Therefore $m^2 + n^2$ is never divisible by $m + n$ .

Relevant wiki: Modular Arithmetic - Problem Solving - Basic

$\begin{aligned}\text{Since m and n are }\text{consecutive }\\ \text{Let, }\\ m&=x\\ n&=x+1\\ \text{We have,}\\ m^2+n^2\pmod{m+n}&\equiv x^2+(x+1)^2\pmod{x+(x+1)}\\ &\equiv 2x^2+2x+1 \pmod {2x+1}\\ &\equiv (2x^2+x)+(x+1) \pmod {2x+1}\\ &\equiv(x+1) \pmod {2x+1}\small\hspace{14mm}\color{#3D99F6}2x^2+x\pmod{2x+1}\equiv 0\\ \text{We have } 0<x+1<2x+1 \\ \implies m^2+n^2 \text{is not divisible by }& m+n \text{ for any pair of consecutive positive integers m,n}\end{aligned}$

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

If $m+n | m^2 + n^2$ , then $m + n | 2mn$ . As they are consecutive, $m + n|mn$ .

$m = 0 [n]$ is never satisfied for m higher than $2$ .

(m+n)^2 = m^2 + n^2 + 2mn. The LHS can be divided by m+n. But m^2 + n^2 cannot be divided by m+n unless m and n are 0.

Let $n=m+1$ , then

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

where $m+1$ is the integer remainder of the division $2m^2+2m+1$ by $2m+1$ . But $m+1 = 0$ only if $m=-1$ , which is not positive.

Lets have m^2 + n^2 modified as (m+n)^2 - 2mn

Hence (m^2 + n^2) / (m+n) is equivalent to (m+n)^2)/(m+n) - 2mn/(m+n)

So the question is just about the last term 2mn/(m+n)

Numerator 2mn - always even Denominator m+n - always odd (m,n are consequtive)

Hence they never divide each other.

I think my solution is a bit lengthy compared to other, but this is what I found.

Suppose there exists 2 consecutive positive integers $m$ and $n$ such that $m+n | m^2 + n^2$

Let $m = k$

$n=k+1$

for some $k \in \mathbb N$

$\Rightarrow$

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

$\Rightarrow$

$m+n = 2k+1$

Then,

$\frac{2k^2 + 2k + 1}{2k+1} = \frac{2k^2}{2k+1} + 1$

Now we have to show that $\frac{2k^2}{2k+1}$ is irreducible $\forall$ $k$ $\in$ $\mathbb N$

By Extended Euclidean Algorithm, one can show that,

$2k^2 = (2k+1)(k) - k$

$\Rightarrow$

$2k^2 - (2k+1)k = -k$

$2k+1 = (-k)(-2) +1$

$\Rightarrow$

$2k+1 + 2(-k) = 1$

Substituting $-k$ in the equation, we have

$2k+1 + 2[2k^2 - (2k=1)k] = 1$

$\Rightarrow$

$(2k+1)(1-2k) + 2(2k^2) = 1$

Therefore there are $\textbf{No}$ consecutive positive integers that satisfy the condition.

My question was to ask what is the result of dividing $m^2 + n^2$ by $m + n$ ? The answer must be less than $n$ because $n * (m + n) = n * m + n * n$ which is more than $m * m + n * n$ . And it must be more than $m$ because $m * (m + n) = m * m + m * n$ which is less than $m * m + n * n$ . And there's no positive integer between two consecutive integers so it will never be exactly divisible, the answer will always be $m$ with remainder $n$ .

Any number squared results always in an even number.

One of two consecutive numbers is always odd. So the sum two consecutive numbers is always odd.

Since even numbers can never be divisible by odd numbers the statement has to be true.

Because $m$ and $n$ are consecutive:

$n=m+1$

In order to be divisible by $m+n$ there must be an integer $k$ that fulfils:

$m^2+(m+1)^2=k·(m+(m+1))$

By reordering we get:

$2m^2+(2-2k)·m+(1-k)=0$

where we can solve $m$ as a second order equation:

$m=\frac { k-1\pm \sqrt { { k }^{ 2 }-1 } }{ 2 }$

We know that $m$ must be an integer, so the square root must solve as an rational number. The only way $\sqrt { { k }^{ 2 }-1 }$ is rational, is when $k=1$ , which gives the solution $m=0$ , $n=1$ , which is not valid because $m$ is not positive. So, there is no valid solution.

◆int x≧3 ◆m=(x-1)/2 ◆n=(x+1)/2 ◆(m²+n²)/(m+n)=(((x-1)/2)²+((x+1)/2)²)/((x-1)/2+(x+1)/2)=…=1/2(x+1/x) ◆m²+n² is never divisible bym+n .

$m^2-n^2$ is divisible by $m+n$

For $m^2+n^2$ to be divisible, $(m^2-n^2)-(m^2+n^2)=k(m+n)$

Or, $2n^2=k(m+n)$

$2n^2=k(2m+1)$ (m and n are consecutive);

The above expression is clearly false as m,n are integers.

Suppose, that the fact is $false$ .

Let those numbers be $k$ and $k+1$ .

Then $k^2 + (k+1)^2 = 2k^2 + 2k + 1 = 1\cdot(k + (k+1)) + 2k^2$ . So, we need to prove that $2k^2$ is divisible by $k + (k+1) = 2k + 1$ .

However, $2k^2$ is always even and $2k + 1$ is always odd. No even number is divisible by odd number.

We got a contradiction, hence the fact is $\boxed{ true }$

Let $m=n+1$ . Therefore, $m^2+n^2=2n^2+2n+1=2n^2+(m+n)$ which is divisible by $m+n$ if and only if $2n^2 = k(m+n)$ , where $k=1,2,3,...$ . In other words $2n^2-2kn-k=0$ . This quadratic eqution solves for $n=\frac{k}{2}\left(1\pm \sqrt{1+\frac{2}{k}}\right)$ , which is not an integer because $1+\frac{2}{k}$ cannot be a square number.