Square the Factorial

The value of $x$ will never lie between 10 to 40

$z! = x^{2} - y^{2}$

$z! = (x - y)(x+y)$

then by Putting $z=5$ we may find that no value of x,y satisfy the equation.

At $z=6$ one case will be possible that is $x=63$ , $y=57$

Similarly at $z=7$ we get the remaining 3 cases

$x=71 , y=1$

$x=72 , y=12$

$x=73 , y=17$

So the answer is 63+57+71+72+73+17+12+1=366

Elaborating @shubhendra singh 's solution:

Since $z!=(x-y)(x+y)$ , we shall be seeking two integers that gives $z!$ when multiplied:

• which add up to an even number (because the sum of the two factors is $(x-y)+(x+y)=2x$ which is even)
• and which lies between $20z$ and $20(z+1)$ (since $x$ lies between $10z$ and $10(z+1)$ ) (assuming that $x$ has only two digits)

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• No such pair of number exists when z is 1 to 5.
• Listing out all the factors of $6!$ gives only $x=63$ as a solution.
• Listing out the factors of $7!$ from its square root gives us three solutions: $x=71$ , $x=72$ and $x=73$ .

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Hence the answer.

z=6: (63,57), z=7: (71,1),(72,12),(73,17) are smallest 4,i calculate almost all. Next are (282,198),(388,332),(687,657),(1686,1674),(10081,10079).For z=9 there is much more solutions...