Find two consecutive positive integers, the sum of whose squares is 365.
Positive integers are 13 and 14.
Step 1: Represent the consecutive integers. Let the smaller positive integer be \(n\). Then the next consecutive integer is \(n + 1\).
Step 2: Use the condition on the sum of squares. The sum of the squares of these two integers is given as 365. So we write:
\[n^2 + (n + 1)^2 = 365\]
Step 3: Expand and simplify.
First expand \((n + 1)^2\):
\[(n + 1)^2 = n^2 + 2n + 1\]
Substitute back into the equation:
\[n^2 + n^2 + 2n + 1 = 365\]
\[2n^2 + 2n + 1 = 365\]
Now move 365 to the left side:
\[2n^2 + 2n + 1 - 365 = 0\]
\[2n^2 + 2n - 364 = 0\]
Step 4: Simplify the quadratic equation. Divide the whole equation by 2 to make it simpler:
\[n^2 + n - 182 = 0\]
Step 5: Factorise the quadratic. We look for two numbers whose product is \(-182\) and whose sum is \(+1\). These numbers are \(+14\) and \(-13\) because:
\[14 \times (-13) = -182, \quad 14 + (-13) = 1\]
So we can factorise as:
\[n^2 + n - 182 = (n - 13)(n + 14) = 0\]
Step 6: Solve for \(n\).
From \(n - 13 = 0\), we get \(n = 13\).
From \(n + 14 = 0\), we get \(n = -14\).
Step 7: Choose the positive solution. The question asks for positive integers, so we discard \(n = -14\) and take:
\[n = 13\]
The next consecutive integer is:
\[n + 1 = 14\]
Conclusion: The two consecutive positive integers are 13 and 14.