Question

find 2 consecutive positive when integers such sun of swires is 884
It's an shape

Answer 1

Let's find two consecutive positive integers whose sum of squares equals 884. We can represent the two consecutive integers as \( n \) and \( n + 1 \).

The equation for the sum of their squares will be:

\[
n^2 + (n + 1)^2 = 884
\]

Expanding the equation, we have:

\[
n^2 + (n^2 + 2n + 1) = 884
\]

This simplifies to:

\[
2n^2 + 2n + 1 = 884
\]

Subtracting 884 from both sides gives:

\[
2n^2 + 2n + 1 - 884 = 0
\]

This simplifies to:

\[
2n^2 + 2n - 883 = 0
\]

We can divide the entire equation by 2 to make it simpler:

\[
n^2 + n - 441.5 = 0
\]

Since this isn't yielding integer results, let's revert to the standard format and solve it as a quadratic equation:

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2 \), \( b = 2 \), and \( c = -883 \):

\[
b^2 - 4ac = 2^2 - 4 \times 2 \times (-883) = 4 + 7064 = 7068
\]

Now let's calculate \( n \):

\[
n = \frac{-2 \pm \sqrt{7068}}{2 \times 2} = \frac{-2 \pm \sqrt{7068}}{4}
\]

Calculating \( \sqrt{7068} \):

The approximate value is \( \sqrt{7068} \approx 84 \).

So,

\[
n \approx \frac{-2 + 84}{4} \approx \frac{82}{4} = 20.5
\]

This indicates integers \( n = 20 \) and \( n = 21 \).

Now confirm their squares:

\[
20^2 + 21^2 = 400 + 441 = 841 \text{ (not equal to 884)}
\]

Let's try another attempt recognizing any misevaluation possibilities.

The correct foundation starts from reevaluating sums or reconsidering integer forms.

If instead, we outlined our integers adequately positioned, \( n^2 + (n+1)^2 \) could yield different breakups repeating through trial and numeral test algorithms:

The achieved \( n \) may be evaluated systematically from bounds set \( n \geq 20, 21 \).

It appears direct pairs yield 30,31 to test proximate sums as posit:

Checking:

\[
30^2 + 31^2 = 900 + 961 = 1861 \quad \text{(again doesn't fit)}
\]

Employ guidance back to consistency of squares, upcasting and retrying \(\pm\) ranges consistently.

Final attempt resolves while relying foundational iterations for indexed integers resolving back \( n \) resumes at:

### Conclude:

Unfortunately, instances derived herein challenge 884 from duplicative integer sets iteratively proven through structured sequences remain.

Each peak vis-a-vis must be re-evaluated for leaving derivational base for congruence; thus failing reflects around improper iterative solutions confirming.

**Answer: Conclusively testing intervals may necessitate acceptance of closer proximity integers not directly yielding the required property mathematically iterated or forward resolved**.

**If needed please request deeper directly indexed boundaries for coherence**.