1^2 = 1 2^2 = 1 + 3 3^2 = 1 + 3 + 5 4^2 = 1 + 3 + 5 + 7 n^2 = 1 + 3 + ... + (2n-1)Here's a picture of how this series works:

1^2 = 1 2^2 = 1 3 3 3 3^2 = 1 3 5 3 3 5 5 5 5 4^2 = 1 3 5 7 3 3 5 7 5 5 5 7 7 7 7 7and so on... Each squared number is the previous square with one more layer which is always two more than the one before it.

To make the following algorithm work well on a calculating machine, Friden used the same series multiplied by 5:

5*(1^2) = 5*(1) = 5 5*(2^2) = 5*(1 + 3) = 5 + 15 5*(3^2) = 5*(1 + 3 + 5) = 5 + 15 + 25 5*(4^2) = 5*(1 + 3 + 5 + 7) = 5 + 15 + 25 + 35 5*(n^2) = 5*(1 + 3 + ... + (2n-1)) = 5 + 15 + ... + (10n-5)

This was more efficient for the machine because it turned multiplications by 20 into multiplications by 100 which can be done by shifting the carriage.

The above is all you need to start brute-forcing square roots by subtracting 1, 3, ... but that would take a very long time for large numbers. Thus you want to proceed from left to right as you did in school.

Begin by assuming we have already found the square root of some of the leading digits of the number n. Call this "root so far" s. (At the first step set s to zero.)

If we square s and subtract it from the original number n, we are left with some remainder r. (i.e. r = n-s^2)

So given an s (root so far), we need to find the next digit d. Appending the next digit would cause a change in the remainder which could be expressed (scaled by 5) as:

5[r1-r2] = 5[n-(10s)^2 - (n-(10s+d)^2)] = 5[-100s^2 + 100s^2 + 20sd + d^2] = 5[20sd+d^2] = 100sd + 5d^2Note that this is also the sum of a (scaled by 5) odd digits series:

100s+5 + 100s+15 ... + 100s+(10d-5)(For example for d=3 we have 100s*3+5*9 = 100s+5 + 100s+15 + 100s+25)

Thus the calculator could compute d by successively subtracting 100s+5, 100s+15 until a negative number would result.

3125Then it would start subtracting 100s+5 100s+15... from the leftmost digits: (s started at 0)

3125 500- 1500- : 2 subtractions so first digit is 2 (so s is now 2) leaving 1125Then it subtracted 100s+5, 100s+15... (s = 2)

1125 205- 215- 225- 235- 245- : 5 subtractions (so d is 5 which is appended to s yielding 25) leaving 0So the root of 625 is 25.

12500 500- 1500- 2500- 3500- 4500- : s is 5 Placing the decimal yields 5 or 50 respectively.

1250 500- : d is 1, s is now 1 750 105- 115- 125- 135- 145- d is 5, s is now 15 12500 1505- 1515- 1525- 1535- 1545- 1555- 1565- 1575- : d is 8, s is now 158 18000 15805- : d is 1, s is now 1581 219500 158105- : d is 1, s is now 15811 6139500 1581105- 1581115- 1581125- : d is 3, s is now 158113And so on... Placing the decimal in the result yields 15.8113 for 250 or 158.113 for 25000. The bottom row of keys on the Friden keyboard indicated the decimal position in the number and started the square root process.

Count Term: Term: Odd Digits Scaled By Form Five Form 1 01 05 2 03 15 3 05 25 4 07 35 5 09 45 6 11 55 7 13 65 8 15 75 9 17 85 10 19 95In the scaled by 5 case, if the 5 in the term is cleared, the digit left in the term is the count minus 1. This was very handy because, instead of doing the shift/add on each iteration to build the new s from the old s and d, the calculator could produce the new s' directly by simply subtracting until the negative number appeared (an overdraft) and then adding that last term back into the remainder.

It then kept this last term (which is one beyond the last term that could be subtracted and thus contained the correct count) and cleared the 5. Since it was much easier for the calculator to detect an actual overdraft than to detect that an overdraft would occur, this algorithm was more efficient all around.

959220 50000- : Now staring with s=0, subtract 05, 15... 150000- 250000- 350000- 450000- : Overdraft so add this last term (450000) back. Clear the 5 keeping 400000 as the root so far. Now start adding 05, 15 to the term (shifted one place to the right) 1592200 405000- : s=4, subtract s05, s15... 415000- 425000- 435000- : Overdraft so add this term (435000) back. Clear the 5 keeping 430000 as the root so far. Add 05, 15 in the next place to the right. 3472000 430500- : s=43, subtract s05, s15... 431500- 432500- 433500- 434500- 435500- 436500- 437500- : equals 0 but no overdraft so continue 438500- : Overdraft so add this term back (leaving 0.) Clear the 5.Since the result is 0, the root is 438000 (438.000) Because of the scaled-by-five algorithm, no adding of d's to s' was required. (The Friden would continue attempting to subtract until it ran out of digits, but we'll stop here since we know we're done.)