Since both variables always have the same value, you actually only need one of them. In Scala, you would generally not use a loop to solve this problem, but use higher-level collection operations instead. Something like:

(0 to 9) map { i => s"i = $i :: j = $i" } mkString "\n"

Note: this will only generate the string that you want to print, but not actually print it. It is generally considered a good thing to not mix generating data and printing data.

If you want to print this, you only need to pass it to println:

println((0 to 9) map { i => s"i = $i :: j = $i" } mkString "\n")

Or, in Scala 2.13+:

import scala.util.chaining._

(0 to 9) map { i => s"i = $i :: j = $i" } mkString "\n" pipe println

You could also write it like this:

(for (i <- 0 to 9) yield s"i = $i :: j = $i") mkString "\n"

Now, you might say, "Wait a minute, didn't you just say that we don't use loops in Scala?" Well, here's the thing: that's not a loop! That is a for comprehension. It is actually syntactic sugar for collection operations.

for (foo <- bar) yield baz(foo)

is actually just syntactic sugar for

bar map { foo => baz(foo) }

A for comprehension simply desugars into calls to map, flatMap, foreach, and withFilter. It is not a loop.

Note that Scala does have a while loop. It exists mainly for performance reasons. Unless you are writing low-level libraries that are going to be used in performance-intensive code by tens of thousands of developers, please just pretend that it doesn't exist.

Also note that if the while loop weren't built into Scala, you could easily write it yourself:

def whiley(cond: => Boolean)(body: => Unit): Unit =
  if (cond) { body; whiley(cond)(body) }

According to http://alvinalexander.com/scala/iterating-scala-lists-foreach-for-comprehension:

val names = Vector("Bob", "Fred", "Joe", "Julia", "Kim")

for (name <- names)
    println(name)