How to prove $xy=n^2$?

Question

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Problem Restatement

We are given two integers (x) and (y).

Their (arithmetic) mean is (n): [ \frac{x+y}{2}=n . ]
The mean of their reciprocals (\dfrac{1}{x}) and (\dfrac{1}{y}) is (\dfrac{1}{n}): [ \frac{\dfrac{1}{x}+\dfrac{1}{y}}{2}= \frac{1}{n}. ]

We have to prove that the product of the two integers equals the square of the mean, i.e.

[ \boxed{\,xy = n^{2}\,}. ]

[ \begin{aligned} \text{(i)}\qquad &\frac{x+y}{2}=n,\[4pt] \text{(ii)}\qquad &\frac{\dfrac{1}{x}+\dfrac{1}{y}}{2}= \frac{1}{n}. \end{aligned} ]

Both equations are valid for the (unknown) integers (x,y) and the (unknown) integer (n).

From (i) multiply both sides by (2):

[ x+y = 2n. \tag{1} ]

From (ii) multiply both sides by (2) and combine the fractions on the left:

[ \frac{1}{x}+\frac{1}{y}= \frac{2}{n}. ]

Put the two fractions over a common denominator (xy):

[ \frac{y+x}{xy}= \frac{2}{n}. \tag{2} ]

Equation (2) contains the same numerator (x+y) that we already know from (1).
Substitute (x+y = 2n) into (2):

[ \frac{2n}{xy}= \frac{2}{n}. ]

Both sides have a factor (2); cancel it (or divide the whole equation by (2)):

[ \frac{n}{xy}= \frac{1}{n}. ]

Now cross‑multiply (multiply each side by (xy) and by (n)):

[ n\cdot n = xy. ]

Thus

[ \boxed{xy = n^{2}}. ]

The required relationship follows directly from the two mean conditions.

The product of the two integers equals the square of their mean:

[ \displaystyle xy = n^{2}. ]

Mistake	Why it’s wrong	How to avoid it
Cancelling the denominator incorrectly (e.g., turning (\frac{y+x}{xy}= \frac{2}{n}) into (y+x = 2) by “cancelling” (xy) and (n)).	You can only cancel a factor that appears both in the numerator and denominator of the same fraction.	Keep the fractions intact until you substitute the known expression for the numerator, then cross‑multiply.
Forgetting to use the first equation after writing the second one.	The second equation still contains (x+y); without replacing it you cannot isolate (xy).	After obtaining (\frac{x+y}{xy}= \frac{2}{n}), immediately replace (x+y) with (2n) from the first equation.
Multiplying by the wrong term (e.g., multiplying (ii) by (n) instead of by (2)).	The original equation has a factor (2) on the left; dropping it changes the equality.	Follow each algebraic step carefully: first clear the outer denominator (multiply by 2), then combine the inner fractions.
Assuming (x) or (y) could be zero.	The reciprocal (\frac{1}{x}) (or (\frac{1}{y})) would be undefined if (x=0) or (y=0).	Remember that the problem implicitly requires (x\neq0) and (y\neq0); the algebra works only under that condition.

Keeping these points in mind will help you solve similar “mean of numbers and mean of reciprocals” problems without errors.

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