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Do you prefer squaring n then multiplying by 25, or is it better to do the squares that end in 5 (so, for 407, do you prefer 400*400 + 100*7*8 + 49 or 2500 * 64 + 100*7*8 + 49)?

That's an excellent question.  That formula was more for proof concept (i.e., eliminating carries and making the units and tens digit controlled only by the y² term).  In practice, I don't pay attention to what the n is nor think of the formula.

Mentally, what I'd do for 407²:

1. Double the 4 is 8
2. Multiply that by 7.  That's 56.
3. 40² is 1600
4. Add them: 1656
5. Multiply by 100: 165,600
6. Put the 7² on the end.
7. 165,649

I do the second term first, because that's the variable and most complicated term...and by complicated, I mean 2 * 4 * 7.

Ones where the first term ends in 50 are more tricky, because the 350² has four-digits instead of just two.  That means I'm going to have to do real addition or subtraction.  For 333²:

1. (350 - 17)²
2. Double the 3.5 is 7
3. Multiply 7 by 17 = 119
4. 35² is 1225
5. 1225 - 119 = 1106
6. Multiply by 100 = 110,600
7. Add 17² (289).
8. 110,889

Step 5 annoys me,  So, taking advantage of the fact that, for practical purposes, I have all two-digit squares memorized, I would do calculate this as (300 + 33)².  Rounding to the nearest 100 usually causes the middle term have more digits  However, the first term now only affects the ten-thousands and hundred-thousands digit, making the adding or subtracting even easier:

1. (300 + 33)²
2. Double the 3 is 6
3. Multiply 6 by 33 = 198
4. 30² is 900
5. 900 + 198 = 1098
6. Multiply by 100 = 109,800
7. Add 33² (1089).
8. 110,889

Step 5 is now trivial, though the addition in step 7 is likely to be more complicated.  It will be an addition and an addition of at most two digits...in this case, ignoring the tens and units digits, the addition is just 1098 + 10.  Using 350², I had to subtract a 3-digit number from a 4-digit number.  The good thing there is that I'm extremely familiar with the 4-digit number and that number will always end in 25.

I am a bit inconsistent when squaring numbers that end in 26 through 74 on whether I round to the nearest 50 vs. the nearest 100.  However, I never use the first term as 450, 550, 950, or 1050.  That's because using 500 and 1000 are even easier, because the middle term will have a 0 as its hundreds digit:

(500 ± y)² = 
250,000 ± 2 * 500 * y + y² =
250 * 1000 ± y * 1000 + y² =
(250 ± y) * 1000 + y² =

That is super easy.  I just add or subtract y from 250.  That's the thousands.  Then add y².  

For 576² -> 250 + 76 = 326 -> 326,000 + 76².

76² = 5776 -> 576² = 331,776

And, in practice, rather than add or subtract y from 250, I take the original 576 and subtract 250 from it.  It equals, 326 either way.

There are several other optimizations from having done this multiple times, but I've already complicated it more than is ideal.  This may be something to bookmark or save somewhere and revisit to incrementally develop your ability, if you want to do it at all.

Even if you don't have squares up to 25² memorized, you could use this for calculating squares between 40² and 60² or 90² and 110² at first and build from there.

I mean we have phones/calculators that can calculate squares for us. To me, memorizing squares/doing mental math is just a fun way to pass the time + a neat trick to be able to do.

It is a stupid human trick and, yes, it would be more useful in the years before people carried computers in their pockets everywhere.  However, being able to quickly do mental arithmetic is still helpful.  It gives you more information than you'd have otherwise.  In an extremely dorky hypothetical, it can allow you to quickly spot when your local Which Wich sub shop is calculating 8.25% incorrectly on your $5.75 sub.  You can then inform them and their district office and they'll give you free subs.  

Hypothetically.