Question

The sum of all natural numbers a such that a2 16a + 67 is a perfect square is:

Answer 1

can be written as

${a}^{2} - 2 \times a \times 8 + {8}^{2} + 3$

$= ({a - 8)}^{2} + 3$

as we can see this expression itself is not a perfect square. hence only way it will be perfect is when (a-8) equals 1 ; to get 1+3=4 which is perfect.

thus the possible natural values are 7 and 9.

so the sum is 7+9=16.