The sum of the squares of three positive integers is 323. If the sum of the squares of two of these integers is equal to twice the third integer, what is the product of the three integers?

Introduction / Context:
This question involves number theory and algebra together. We are told about three positive integers whose squares satisfy a special relationship. The sum of their squares is fixed, and the sum of the squares of two of the numbers equals twice the third number. The objective is to find the product of the three integers. Problems like this appear in aptitude and olympiad style exams to test comfort with manipulating equations and trying small integer possibilities logically.

Given Data / Assumptions:

• Three positive integers a, b, and c.
• a^2 + b^2 + c^2 = 323.
• The sum of the squares of two numbers is equal to twice the third number, which can be written as a^2 + b^2 = 2c, or similar, for some choice of which is the third.
• All three numbers are positive integers.

Concept / Approach:
The total sum of squares 323 is not very large, so the integers themselves must also be relatively small. A practical strategy is to assume that the linear relation a^2 + b^2 = 2c holds for some ordering of the numbers and test reasonable values. Because 2c must equal a^2 + b^2, c will be roughly half of a^2 + b^2, which guides the trial. Once a valid triple is found that satisfies both the sum of squares and the linear condition, we can compute their product and match it to the options.

Step-by-Step Solution:
Assume three positive integers a, b, c such that a^2 + b^2 + c^2 = 323.We are told that the sum of squares of two numbers equals twice the third. A suitable interpretation is a^2 + b^2 = 2c.Then c must be an integer, so a^2 + b^2 must be even.Try small integer values where squares add up near 323 when combined with a third square.Check a = 3, b = 5: then a^2 + b^2 = 9 + 25 = 34.If a^2 + b^2 = 2c, then 34 = 2c which gives c = 17, an integer.Now check the sum of squares: 3^2 + 5^2 + 17^2 = 9 + 25 + 289 = 323, which matches the given total.Therefore, the three integers are 3, 5, and 17.Their product is 3 * 5 * 17 = 255.

Verification / Alternative check:
Verify the special relation: a^2 + b^2 = 34 and 2c = 2 * 17 = 34, so the condition a^2 + b^2 = 2c holds.Verify the sum of squares again: 9 + 25 + 289 = 323, exactly as required.Thus the triple (3, 5, 17) satisfies all the conditions and its product 255 is consistent.

Why Other Options Are Wrong:
260, 265, and 270 correspond to other hypothetical products, but no other triple of positive integers satisfies both the sum of squares condition and the special relation a^2 + b^2 = 2c.Any different combination that gives these products either violates the sum of squares 323 or fails to satisfy the relation between the two squares and twice the third.

Common Pitfalls:
One common mistake is to misread twice the third as twice the square of the third, which would change the equation to a^2 + b^2 = 2c^2 and lead to no valid integer solutions in this case.Another error is to assume that the three numbers must be in ascending order and then restrict trials too much, missing the correct combination.Some learners also forget to check both the relation and the sum of squares, verifying only one of the two conditions.

Final Answer:
The product of the three positive integers is 255.