Squares and Sums I

Here's an approach to the problem that is a little more rigorous than trial and error:

The key to solving this problem is a familiarity with the concept of the difference of two squares .

Let $x+38=a^2$ and let $x+19=b^2$ for positive integers $a$ and $b$ ( $a>b$ )

So $a^2-b^2=19$ , which we can rewrite as $(a+b)(a-b)=19$

Considering the positive factors of 19, we have $(1)(19)$ only. So now we can set up a pair of simultaneous equations:

$a+b=19$

$a-b=1$

The solutions to these are $a=10, b=9$

Now we can sub one of these values back into one of our original equations involving $x$ (let's use $a$ for now, although $b$ would yield an equally valid solution):

$x+38=(10)^2=100$

$x=62$

And sure enough, $62+38=100, 62+19=81$

Now since there is only one pair of positive factors for 19 (as it is prime), there is only one solution for $x$ .

62 is the only value for $x$ which satisfies the conditions so the sum of all the integers $x$ is simply 62 .

Here's a similar problem to try: Squares and sums II

let $x+38=a^2..........[1]$ and $x+19=b^2..........[2]$ where $a,b \in \mathbb{N}$

$[1]-[2]\Rightarrow \cancel{x}+38-\cancel{x}-19=a^2-b^2$ $\Rightarrow 19=(a-b)(a+b)$

$Case 1$ : $a-b=1..........[A_1]$ $a+b=19..........[B_1]$ $[A_1]+[B_1]\Rightarrow 2a=20 \therefore \boxed{a=10}$ $\therefore x+38=10^2=100\Rightarrow\boxed{x=62}$

$Case 2$ : $a-b=19..........[A_2]$ $a+b=1..........[B_2]$ $[A_2]+[B_2]\Rightarrow 2a=20 \therefore \boxed{a=10}$ $\therefore x+38=10^2=100\Rightarrow\boxed{x=62}$

$Case 3$ : $a-b=-19..........[A_3]$ $a+b=-1..........[B_3]$ $[A_3]+[B_3]\Rightarrow 2a=-20 \therefore \boxed{a=-10}$ $\therefore x+38=(-10)^2=100\Rightarrow\boxed{x=62}$

Therefore the only possible value of $x$ is $\boxed{\blue{62}}$