There is a natural number which becomes equal to the square of a natural number when $100$ is added to it, and to the square of another natural number when $168$ is added to it. Find the number.

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Solution

From the question, we can write that
$a$ + $100$ = $b^{2}$
$a$ + $168$ = $c^{2}$
=> $c^{2} - b^{2}$ = 68
=> $(c - b) (c + b) = 1 \times 68$
Now solve the equation for $(c - b) = 1$ and $(c + b) = 68$
We get $c = 35$ and $b = 34$ . (here $c = 34.5$ which is $35$ and $b = 33.5$ which can be written as $34$ by ignoring the point )
$a + 100 = b^{2}$
$a = 34^{2} - 100$
$a = 1156 - 100 = 1056$ hence this the required number.

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There is a natural number which becomes equal to the square of a natural number when 100 is added to it, and to the square of another natural number when 168 is added to it. Find the number.

There is a natural number which becomes equal to the square of a natural number when $100$ is added to it, and to the square of another natural number when $168$ is added to it. Find the number.