The English mathematician Augustus DeMorgan, who lived in the 19th century, once remarked that he was $x$ years old in the year x^2. When was he born?

Question

I found this from Elementary Number theory from Koshy. The answer is $1806$, but does know tell how to find? Anyone have any idea how to solve this problem ?

Answer 1

Assume the solution is a whole number.

Look at squares in between 1800 and 1900. As $40^2=1600$ trying the next few numbers shows $43^2=1849$ is in that range. That shows he must be born in year $1849-43=1806$.

Answer 2

Living in the $19^{th}$ century means born somewhere between $1700$ and $1900$ (with a large safety margin, more than $100$ years lifetime). This corresponds to possible $x$ in range $\left\lceil\sqrt{1700\frac14}+\frac12\right\rceil=42$ to $\left\lfloor\sqrt{1900\frac14}+\frac12\right\rfloor=44$ inclusive.

$$\color{green}{42^2-42=1722,\\43^2-43=1806,\\44^2-44=1892}.$$

All three solutions are possible! As written, the problem statement doesn't really allow to reject $1722$, though he would have been aged $42$ in the $17^{th}$ century and needed to live at least $79$ years to reach the $19^{th}$, nor $1892$, though he would have been aged $44$ in the $20^{th}$ century (but still born in the $19^{th}$).

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A stronger interpretation, "lived only in the $19^{th}$ century" (which happens to be true) restricts the range to $43,44$, and allows to rule out $1892$, as that birth year would imply that he died before the age of $9$.

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The strange formulas come from

$$a\le x^2-x\le b$$ $$a+\frac14\le\left(x-\frac12\right)^2\le b+\frac14$$ $$\sqrt{a+\frac14}+\frac12\le x\le\sqrt{b+\frac14}+\frac12$$