Squares and Sums II

See my solution on Squares and sums for a head start!

Let's use the difference of two squares to solve this one.

We can set up two simultaneous equations:

$x+456=a^2$

$x+123=b^2$

Subtracting these gives $a^2-b^2=333$ , which we can rewrite as $(a+b)(a-b)=333$

We can quickly find that the only positive factors of $333$ are, in pairs, ( $1)(333)$ , $(3)(111)$ and $(9)(37)$

Note: since both $a$ and $b$ will ultimately be squared, we need only consider positive values, and hence the positive factors.

Now we can write:

CASE 1: $a+b=333, a-b=1$ which gives $a=167, b=166$

CASE 2: $a+b=111, a-b=3$ which gives $a=57, b=54$

CASE 3: $a+b=37, a-b=9$ which gives $a=23, b=14$

Substituting these pairs of $a$ and/or $b$ into our two simultaneous equations yields $x=27433, 2793, 73$

Let's check these values:

$27433+456=27889=167^2, 27433+123=27556=166^2$

$2793+456=3249=57^2, 2793+123=2916=54^2$

$73+456=529=23^2, 73+123=196=14^2$

Summing these values for $x$ gives $30299$ .