Sum \$\text{Square}^2\$ Let \$n=42\$ (Input)

Then divisors are : 1, 2, 3, 6, 7, 14, 21, 42

Squaring each divisor : 1, 4, 9, 36, 49, 196, 441, 1764

Taking sum (adding) : 2500

Since \$50\times 50=2500\$ therefore we return a truthy value. If it is not a perfect square, return a falsy value.

Examples :

42  ---&gt; true
1   ---&gt; true
246 ---&gt; true
10  ---&gt; false
16  ---&gt; false


This is code-golf so shortest code in bytes for each language wins 

Thanks to @Arnauld for pointing out the sequence : A046655
R, 39 37 bytes



!sum((y=1:(x=scan()))[!x%%y]^2)^.5%%1


Try it online!

Uses the classic "test if perfect square" approach, taking the non-integral part of the square root S^.5%%1 and taking the logical negation of it, as it maps zero (perfect square) to TRUE and nonzero to FALSE.

Thanks to Robert S for saving a couple of bytes!