Question

Find number of triplets of integers in arithmetic progression the sum of whose squares is $1994$.

Answer 1

Let the middle term be $a$.
Since the numbers are consecutive, hence the numbers will be
$(a-1),a,(a+1)$
Now
$(a-1{)}^2+a^2+(a+1{)}^2=1994$
$a^2-2a+1+a^2+a^2+2a+1=1994$
$3a^2+2=1994$
$3a^2=1992$
$a^2=664$
Now $664$ is not a perfect square.
Hence $a$ is not an integer.
Therefore there are no triplet consecutive integers whose sum of squares is 1994.