Numbers only

1)The sum of the digits of the number is equal to the square of the digit at unit's place. a+b=b^2

2)The product of the digits is equal to double the square of the digit at unit's place. ab=2b^2

find the value of a using the first rule a+b=b^2

a=b^2-b

then find the value of b using the second rule

ab=2b^2

substitute a using the value of a in the first rule

(b^2-b)b=2b^2

b^3-b^2=2b^2

b^3-b^2-2b^2=0

b^3-3b^2=0

b^2(b-3)=0

b-3=0

b=3

using first or second rule find the value of a

a+b=b^2

a+3=3^2

a=9-3

a=6

or

ab=2b^2

a(3)=2(3^2)

a=18/3

a=6

the two digit number is 63

$a+b=b^2$ and $ab=2b^2$ . The second equation tells us $b(a-2b)=0$ . Considering that b is nonzero, $a=2b$ . Now $2b+b=b^2$ which means $2+1=b$ or $b=3$ . Then $a=2(3)=6$ . Thus our number is $\boxed{63}$ .