Question 164607
Let H be the hundred digit.
Then the tens digit is 4H
The ones digit is 4H+1

The "number is (H*100)+(4H*10)+(4H+1) = (100H)+(40H)+(4H+1)= 144H + 1
The sum of the digits is H+4H+4H+1 = 9H+1

You are told the square of the sum of the digits is 72 more than the number itself
So
{{{(9H+1)^2 - 72 = 144H + 1}}}
{{{81H^2 + 18H -71 = 144H + 1}}}
{{{81H^2 -126H - 72 = 0}}}}
{{{(9H + 4)(9H - 18) = 0 }}}
H = -4/9 or 18/9=2
Since the number is an integer, the answer must be H=4

Test your answer. Does {{{289 = 19^2 - 72 }}}