Question 796205
You can try to narrow your choices looking at issues of divisibility and/or modular arithmetic.
Alternatively, there is the straight algebra approach.
I have a feeling that there is a much simpler solution out there, but I will co with my complicated one, because that's all I could think of.
 
STRAIGHT ALGEBRA:
{{{a}}}= tens digit ({{{1<=a<=9}}})
{{{b}}}= ones digit {{{0<=b<=9}}}
{{{10a+b}}}= value of the number
"A two digit number is 18 less than the sum of the squares of its digits" translates as
{{{10a+b=a^2+b^2-18}}}
Here comes the algebra.
{{{10a+b=a^2+b^2-18}}}
{{{a^2+b^2-10a-b=18}}}
{{{a^2-10a+25+b^2-b+1/4=18+25+1/4}}}
{{{(a-5)^2+(b-1/2)^2=173/4}}}
{{{4(a-5)^2+4(b-1/2)^2=173}}}
{{{highlight((2a-10)^2+(2b-1)^2=173)}}}
Since {{{0<=b<=9}}} , {{{0<=2b<=18}}}, and {{{highlight(-1<=2b-1<=17)}}}
Since {{{1<=a<=9}}} , {{{2<=2a<=18}}}, and {{{2-10=highlight(-8<=2a-10<8)=18-10}}}.
There are only 5 possible values for {{{(2a-10)^2}}}, and 4 of then do not lead to a solution.
{{{abs(2a-10)=0}}}-->{{{(2a-10)^2=0}}}-->{{{(2b-1)^2=173}}} has no solution with integer {{{b}}} such that {{{0<=b<=9}}}
{{{abs(2a-10)=4}}}-->{{{(2a-10)^2=16}}}-->{{{16+(2b-1)^2=173}}}-->{{{(2b-1)^2=173-16}}}-->{{{(2b-1)^2=157}}} has no solution with integer {{{b}}} such that {{{0<=b<=9}}}
{{{abs(2a-10)=6}}}-->{{{(2a-10)^2=36}}}-->{{{36+(2b-1)^2=173}}}-->{{{(2b-1)^2=173-36}}}-->{{{(2b-1)^2=137}}} has no solution with integer {{{b}}} such that {{{0<=b<=9}}}
{{{abs(2a-10)=8}}}-->{{{(2a-10)^2=64}}}-->{{{64+(2b-1)^2=173}}}-->{{{(2b-1)^2=173-64}}}-->{{{(2b-1)^2=13709}}} has no solution with integer {{{b}}} such that {{{0<=b<=9}}}


The only solution comes from {{{abs(2a-10)=2}}}:
{{{abs(2a-10)=2}}}-->{{{(2a-10)^2=4}}}-->{{{4+(2b-1)^2=173}}}-->{{{(2b-1)^2=169}}}-->{{{abs(2b-1)=13}}}
but as {{{-1<=2b-1<=17}}}, the only solution requires {{{2b-1=13}}}--> {{{2b=14}}}-->{{{b=14/2}}}-->{{{highlight(b=7)}}}
In turn,
{{{abs(2a-10)=2}}}-->{{{system(2a-10=-2,"or",2a-10=2)}}}-->{{{system(2a=8,"or",2a=12)}}}-->{{{system(highlight(a=4),"or",highlight(a=6))}}}
So the two-digit number is {{{highlight(47)}}} or {{{highlight(67)}}}.


CONSIDERING DIVISIBILITY:
{{{18=2*9}}}
Let's consider divisibility by 2 and by 9.
 
The number could be even, with both digits being even, making their squares, the sums of the squares, and the sum of the squares minus 18 even.
It cannot be even with an odd tens digit, because that would make that digit's square, the sum of the squares, and the sum of the squares minus 18 odd.
It cannot be odd with both digits being odd, because each square would be odd, making their sum even, and the sum of the squares minus 18 even.
The number could be odd if the tens digit is even.
Divisibility by 2, requires that the number be even with two even digits, or else odd, with an even first (tens) digit. Either way, the first digit must be even.

 
If a number is divisible by 9, when divided by 9, the remainder is zero, and the number can be written as {{{9k}}} for some non-negative integer {{{k}}}, as in
{{{0^2=0}}}, {{{3^2=9}}}, {{{6^2=36}}}, and {{{9^2=81}}}.
In that case the number is called congruent with 0 modulo 9, and all those numbers congruent with 0 are called congruent with one another.
 
Other numbers have a remainder of 1, 2, 3, ...., or 8, when divided by 9, and are called congruent modulo 9 with that remainder. 
All the numbers with the same remainder and congruent modulo 9 with one another and with that remainder.
An easy way to find that remainder is to add the digits of the number. If the result is 10 or more, you keep adding the digits of the result, until you get a one digit number. if it is 9, then the remainder is zero. Otherwise, the final result is your remainder. So, if the digits are {{{x}}} and {{{y}}}, the number is congruent with {{{x+y}}} modulo 9.
 
Since their difference is {{{18=2*9}}}, congruent modulo 9 with zero,
the sum of the squares, and the sum of the digits must be congruent modulo 9.
The square of a digit minus the digit itself, {{{x^2-x=x(x-1)}}} can only be congruent with 0, 2, 3, or 6.
Here are the remainders for x=0 through 9:
{{{matrix(3,1,x,x-1,x(x-1))}}}{{{matrix(3,5,0,1,2,3,4,8,0,1,2,3,0,0,2,6,3)}}}{{{matrix(3,5,5,6,7,8,0,4,5,6,7,8,2,3,6,2,0)}}}

If you add those differences for the two digits, we should get 0 or 9.
The only choices are {{{0+0=0}}} and {{{3+6=9}}}
{{{0+0}}} can only be obtained for 10, 11, 19, 90, 91, 99, but all those numbers fail the first requirement we found based on divisibility by 2.
 
To make {{{3+6=9}}},
{{{x(x-1)}}} is congruent with 6 only for the odd digits {{{3}}} and {{{7}}}, which can only be units digits.
{{{x(x-1)}}} is congruent with 3 only for the even digits {{{4}}} and {{{6}}}, which will have to be tens digits.
The only choices left are 43, 47, 63, and 67.
{{{43<>4^2+3^2-18=16+9-18=7}}} so 43 is not a solution
{{{63<>6^2+3^2-18=36+9-18=27}}} so 63 is not a solution
{{{47=4^2+7^2-18=16+49-18=highlight(47)}}}
{{{67=6^2+7^2-18=36+49-18=highlight(67)}}}