**Tell us what’s happening:**

Can someone explain me this code line:

```
const squaredIntegers = arr.filter( (num) => num > 0 && num % parseInt(num) === 0).map ( (num) => Math.pow(num,2) );
```

I dont understand why the solution have `&& num % parseInt(num)===0`

and the function `(num) =>Math.pow(num,2)`

What does Math.pow doing? and the if case `&& num % parseINt(num)===0`

is everytime time true or not? example with integer 4.2: parseInt(4.2) makes from 4.2 into a Int 4. and 4 % 4 === 0 is true. So by any number, dont matter which number i take, the solution in this part is always true. so why we have this if case here?

Thanks and sorry for my bad english

**Your code so far**

```
const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2];
const squareList = (arr) => {
"use strict";
// change code below this line
const squaredIntegers = arr.filter( (num) => num > 0 && num % parseInt(num) === 0).map ( (num) => Math.pow(num,2) );
// change code above this line
return squaredIntegers;
};
// test your code
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);
```

**Your browser information:**

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.

**Link to the challenge:**

https://learn.freecodecamp.org/javascript-algorithms-and-data-structures/es6/write-higher-order-arrow-functions