Square Root of MATHS challenge problem

[Bob Brown MSEE]

Let S_n be an infinite sequence of integers. The first 7 elements are listed below.
S₁ = 11
S₂ = 817
S₃ = 5622
S₄ = 94995
S₅ = 11177247
S₆ = 755535173
S₇ = 13180586707
S₈ = ?
S₉ = ?
S₁₀ = 436942156724195

Challenge:​
1) Find values for S₈ and S₉
2) Is S₁₀ the correct value?
3) If not, find the correct S₁₀ value.

Note:
Whether or not S₁₀ is the correct value, S₁₀ is an extremely important clue. You can click on the value of S₁₀ for more information.

 

[Bob Brown MSEE]

Been hitting the jug, Bob?

Hi Dennis, ha, sounds like it -- I agree.
However, I've learned a great deal about p-adic numbers recently and designed a very similar version of this challenge for my son. This problem is a painless way to get a LOT of insight.

What I like about this problem is that it is very simple. It is well within the grasp of advanced middle-school students. However, it seems completely insane!

 

[Bob Brown MSEE]

Hints:
1) Make a list of BOTH S_n and S_n² in base 31.
2) Notice that S_n² improves by exactly one digit at each step.
i.e. There are n significate digits in S_n²
3) What do you notice about S_n in base 31?
4) Try to predict S₈ and S₉ in base 31.

Comments:
This type of problem is a typical, "find the next number in this list".
I am not a fan of this type of problem, because there are usually an infinite number of solutions. This problem is not quite that trivial. The extra info like "Square root of MATHS" and what you see when you click S₁₀ provides sufficient constraint to uniquely solve the Challenge.

 

[Janis]

Hi Bob,

Hints:
1) Make a list of BOTH S_n and S_n² in base 31.
2) Notice that S_n² improves by exactly one digit at each step.
i.e. There are n significate digits in S_n²
3) What do you notice about S_n in base 31?
4) Try to predict S₈ and S₉ in base 31.

Ok, I followed your hint and I get...

n	Sn	Sn2
1	b31	3s31
2	qb31​	mchs31
3	5qb31​	136ths31
4	35qb31​	a56aths31
5	c35qb31​	4meqhmaths31
6	qc35qb31​	mee2920maths31
7	eqc35qb31	73hcab300maths31
8	?	?
9	?	?
10	gg9g4irp4k31	8p3cias9d300000maths31

Hints Answers:
1) done (above)
2) improves if rounded by mod 31ⁿ
3) Trailing digts match S_n-1 except S₁₀
It confuses me that S₁₀ does not fit the pattern.
4) Here's my guesses, but don't match S₁ through S_​10

S₈ = 9g4irp4k₃₁
S₉ = g9g4irp4k₃₁

Are there many integer sequences that approach the "square root of maths₃₁" in this way?

 

[Bob Brown MSEE]

Hard way to solution

4) Here's my guesses, but don't match S₁ through S_​10 <-- you are correct

S₈ = 9g4irp4k₃₁ <-- wrong, not in this seqence
S₉ = g9g4irp4k₃₁ <-- wrong, not in this seqence
Are there many integer sequences that approach the "square root of maths₃₁" in this way?<-- No

Very good!

There are no other integer sequences that approach the "square root of maths₃₁" in this way.
We have found two.
The Fundamental theorem of algebra assures that there are only two square roots of an integer.

You are very close to finding S₈ and S₉ of the posted challenge.
The following extension of your table might be useful.




The function IntegerExponent[number, 31] can be used in WolframAlpha.
Using this "IntegerExponent", this can be used to generate both square roots from scratch.
That is how I created this challenge.


EASIER SOLUTION
You have also noticed a shortcut to the solution using S₁₀.
I will explore that in the next section, "Shortcut to Solution"

 

[Janis]

I got it!

S₈ = 590945483038= leqc35qb₃₁
S₉ = 12531420007212= eleqc35qbk₃₁
S₁₀ = 382686130256606= eeleqc35qbk₃₁


You can click each answer to see how I got the first digit.
I also calculated the two square-roots of my name, for extra credit.


First square-root of janis₃₁
...58491141365789 = ...26hu7bnmbk
Second square-root of janis₃₁
...761137145615012 = ...sod0nj78jb₃₁

 

[Bob Brown MSEE]

Shortcut to Solution

4) Here's my guesses, but don't match S₁ through S_​10

S₈ = 9g4irp4k₃₁
S₉ = g9g4irp4k₃₁

You have discovered that S₁₀ is the other square root. Let's call the sequence for the second square root by S2, and the original problem S1.
You also calculated S2₈ and S2₉ (see quote above). This leads to an easy way to solve the challenge using the value of S2₁₀.

You calculated S_n-1 from S_n in base 31 by dropping one leading digit. To emphasize how important your calculation is, lets build your "hints" table for BOTH roots and use your calculation to populate the entire table for S2.

n	S1n	S1n2	n	S2n	S2n2
1	b31	3s31	1	k31	cs31
2	qb31​	mchs31	2	4k31	lhs31
3	5qb31​	136ths31	3	p4k31	kcfths31
4	35qb31​	a56aths31	4	rp4k31	otefaths31
5	c35qb31​	4meqhmaths31	5	irp4k31	bg34qmaths31
6	qc35qb31​	mee2920maths31	6	4irp4k31	l7lib0maths31
7	eqc35qb31	73hcab300maths31	7	g4irp4k31	8co5tkc00maths31
8	?	?	8	9g4irp4k31	2sk2nahb000maths31
9	?	?	9	g9g4irp4k31	8hsicie9b0000maths31
10	?	?	10	gg9g4irp4k31	8p3cias9d300000maths31

Notice the following: for each n in [1,2,..,7]
1) There are n digits in S
2) There are n significant digits in S²
Question: What is the sum of corresponding digits in S1_n and S2_n.
Complete S1₈, S1₉, and S1₁₀, using your answers to this question.