Math Challenge by QuantumQuest #2

[Greg Bernhardt]

Submitted and judged by: @QuantumQuest
Solution credit: @MAGNIBORO

RULES:
1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.

CHALLENGE:
Find all integers $n$ for which the number $[\frac{1^2 + 2^2 +\cdots+n^2}{n}]^\frac{1}{2}$ is integer.

 

[MAGNIBORO]

$$\sqrt{\frac{\sum_{a=1}^{n}a^2}{n}}=N, \, \, \, \, \, \, \, n,N\in \mathbb{N}$$
$$\sqrt{\frac{(2n+1)(n+1)}{6}}=N$$
$$2n^2+3n+1-6N^2=0$$
so
$$n=\frac{-3\pm \sqrt{48N^2+1}}{4}$$
taking the positive root and comparing
$$4n+3 = \sqrt{48N^2+1}$$
because $4n+3$ is a integer , the right hand side Must be an integer $E$
then:
$$ 48N^2 - E^2 = -1 $$
this is a Diophantine equation
$$ Ax^2-y^2=-1,\, \, \, \, \, \, \, \, \, \sqrt{1+A}\in \mathbb{N} $$
with the solutions
$$ x=\frac{\left ( \sqrt{1+A}+\sqrt{A} \right )^m -\left ( \sqrt{1+A}-\sqrt{A} \right )^m}{2\sqrt{A}}$$
$$y=\frac{\left ( \sqrt{1+A}+\sqrt{A} \right )^m +\left ( \sqrt{1+A}-\sqrt{A} \right )^m}{2}$$
for every integer $m > 0$
then
$$E=\frac{\left ( 7+\sqrt{48} \right )^m +\left (7-\sqrt{48} \right )^m}{2}$$
so
$$4n+3= \frac{\left ( 7+\sqrt{48} \right )^m +\left (7-\sqrt{48} \right )^m}{2} $$
finally
$$n= \frac{\left ( 7+\sqrt{48} \right )^m +\left (7-\sqrt{48} \right )^m}{8}-\frac{3}{4} $$
for every integer $m > 0$ that complies with the property that the right side of the equation is an integer
$m=1,3,5,7,...$
(I suspect that they are all the odd integers)

 

[Buzz Bloom]

Hi @MAGNIBORO:

I have a question (Q) and a partial confirmation (P).

Q: Can you please give a reference for your Diophantine equation solutions

x = ...​

and

y = ... .​

I have only a slight education regarding Diophantine equations, and I would appreciate a bit more.

P: Since I was not sufficiently familiar with Diophantine equation methods to confirm your proof, I decided to check a special case of your final equation with the slightly modified form

n = (1/8) × ( (a+b)^m + (a-b)^m ) - 6​

for m = 3, with a = 7 and b = √48 .

U = (a+b)³ = a³ + 3a²b + 3ab² + b³
V = (a-b)³ = a³ - 3a²b + 3ab² - b³
U+V = 2a³ + 6ab² = 2a (a² +3b²)
U+V = 14 × (49 + 3 × 48) = 2702
n = (1/8) × (2702 - 6) = 1696/8 = 337.

I also did a spread sheet for n to 300,000, but I failed to find any solutions for n > 1. I will now need to look for my error there.ADDED
I just noticed as mistake I made with my U and V. I will do some more work.

Regards,
Buzz

 

[QuantumQuest]

@MAGNIBORO

Very well done!

 

[MAGNIBORO]

hi @Buzz Bloom:
to try to proof x=... and y=... let's start with
$$ Ax^2-y^2=-1$$

if $a,b\in \mathbb{R}$
$$a^2-b^2=1$$
then
$$\left ( \frac{\left ( a+b \right )^m-\left ( a-b \right )^m}{2} \right )^2 - \left ( \frac{\left ( a+b \right )^m+\left ( a-b \right )^m}{2} \right )^2= -1$$
by comparison
$$x=\frac{\left ( a+b \right )^m-\left ( a-b \right )^m}{2\sqrt{A}} $$
$$y= \frac{\left ( a+b \right )^m+\left ( a-b \right )^m}{2} $$

using binomial theorem:
$$x=\frac{1}{\sqrt{A}}\sum_{k=1,3,..}^{m}\binom{m}{k}a^{m-k}\, b^k$$
$$y=\sum_{k=0,2,..}^{m}\binom{m}{k}a^{m-k}\, b^k$$

because x and y are integers , the term $\binom{m}{k}a^{m-k}\, b^k$ must be a integer for even values,
and the term $\frac{\binom{m}{k}a^{m-k}\, b^k}{\sqrt{A}}$ must be a integer for odd values

so:

$\binom{m}{k}$ is integer for all k,

$a,a^2,a^3,...$ have to be integers, so $a$ must be integer

$\frac{ b}{\sqrt{A}},\frac{ b^3}{\sqrt{A}},...$ have to be integers

and

$b^0,b^2,b^4,..$ have to be integers

so $b$ must be $\sqrt{A}$

because $a^2 -b^2=1$

$a=\sqrt{1+A}$

and $a$ must be integer so $\sqrt{1+A}\in \mathbb{N}$

finally

$$x=\frac{\left ( \sqrt{1+A}+\sqrt{A} \right )^m -\left ( \sqrt{1+A}-\sqrt{A} \right )^m}{2\sqrt{A}}$$
$$y=\frac{\left ( \sqrt{1+A}+\sqrt{A} \right )^m +\left ( \sqrt{1+A}-\sqrt{A} \right )^m}{2}$$

 

[Buzz Bloom]

Hi @MAGNIBORO:

Thanks for your post. It will take me a while to grasp it.

I have attached below the upper part of my spreadsheet. It shows that for

1 < N < 41,
Col B = (1/N) sum[k = 1 to N] (k^2)​

is not a perfect square.
I am unable to find any flaws in the calculations.

In column E, a zero value means the sum is a square.

Regards,
Buzz

 

[MAGNIBORO]

Hi @MAGNIBORO:

Thanks for your post. It will take me a while to grasp it.

I have attached below the upper part of my spreadsheet. It shows that for

1 < N < 41,
Col B = (1/N) sum[k = 1 to N] (k^2)​

is not a perfect square.
I am unable to find any flaws in the calculations.

View attachment 138820
In columns E, a zero value means the sum is a square.

Regards,
Buzz

I also tried to make a program (in C) but only work for the first 4 or 5 solutions (1, 337,...) then it Lost the precision and mark wrong solutions =(

 

[mfb]

I am unable to find any flaws in the calculations.

What does it do for n=337?

The first few solutions:
n=1
n=337
n=65521
n=12710881
n=2465845537

The numbers grow quickly.

Sequence: A084231

 

[Buzz Bloom]

What does it do for n=337?

Hi @mfb:

(n+1) = 338
(2n+1) = 675
(n+1)(2n+1)/6 = 38025
sqrt(38025) = 195​

Thanks for pointing my error out to me. It wasn't in the calculation but in my search for rows with "OK"

I also found n = 65521 produced
(n+1)(2n+1)/6 = 1431033241
sqrt(1431033241) = 37829

For n < 300001 there were no other solutions.

Regards,
Buzz

 

[Buzz Bloom]

√Hi @MAGNIBORO:

Using the solution at the bottom of your post #2 and letting m = 3, I get that

n = 589.​

This is not a solution of the original problem since

(n+1)(2n+1)/6 = 115,935​

and this is not a perfect square.

Also, n = 337 is a solution of the original problem, and it doesn't seem likely that a value of m will produce this value of n, since n(m) seems to be monotonically increasing with m.

ADDED
Sorry, my bad. I found a bug in the spreadsheet i used to make the calculation. When I fixed the bug I get good answers.
let f(n) = (n+1)(2n+1)/6

m = 1 yields: n = 1, f(n) = 1, √f(n) = 1
m = 3 yields: n = 337, f(n) = 38025, √f(n) = 195
m = 5 yields: n = 65521, f(n) = 1431033241, √f(n) = 37829​

Regards,
Buzz

 

[Buzz Bloom]

(I suspect that they are all the odd integers)

I interpret @MAGNIBORO's remark to indicate a bit of uncertainty about whether or not all odd values of m produce a solution, that is, produces an n which is an integer.
The following is my effort to prove that this is so.

Equation (1) is a slightly edited form of the last equation in Post #2. Each value of m produces a value for n.
Equation (2) is derived from (1) by expanding the two binomial to power m expressions, and then simplifying a bit. Note that when k is odd, the resulting term is zero.

What is both necessary and sufficient in order to prove n is an integer is that S, the summation expression in (2), satisfies

S = 3 mod 4.​

Then -3 + S = 0 mod 4, so the numerator has no remainder when divided by 4, and thus n is an integer.

First look at the term for k = 0. Since m is an odd integer,

S₀ = 7^m = 7 × 49^(m-1)/2.​

Therefore since 7 mod 4 = 3, and 49 = 1 mod 4,

S₀ mod 4 = (7 mod 4) × (49 mod 4)^(m-1)/2 = 3 mod 4.
​

Each of the other terms of S, S_k, k even and k>0, have a factor 48^k/2.
Since 48 = 0 mod 4,

S_k mod 4 = 0.​

Therefore
S mod 4 = 0.

QED

 

[Buzz Bloom]

I have been thinking some more about the "proof" of the correctness of the solution presented in this thread for the OP problem. After some further reflection it occurs to me that the "proof" is incomplete.

We have the following two equations from post #2:

(a) 48 N² - E² = -1,​

where N is the square root of (1/n) times the sum of the squares of the integers
from 1 to n, and

(b) E = 4n+3.
​

Combining these produces

(c) 48 N² - (4n+3)² = -1.​

What has been proved is that for a particular function F(48,m), the equation

(d) n = F(48,m) (as given for n = ... in post #11)​

calculates values of n for which (c) is satisfied if and only if m is an odd integer.

What has NOT yet been proved is that no value of n exists which satisfies (c) other than those calculated from (d).

 

[MAGNIBORO]

I have been thinking some more about the "proof" of the correctness of the solution presented in this thread for the OP problem. After some further reflection it occurs to me that the "proof" is incomplete.

We have the following two equations from post #2:

(a) 48 N² - E² = -1,​

where N is the square root of (1/n) times the sum of the squares of the integers
from 1 to n, and

(b) E = 4n+3.
​

Combining these produces

(c) 48 N² - (4n+3)² = -1.​

What has been proved is that for a particular function F(48,m), the equation

(d) n = F(48,m) (as given for n = ... in post #11)​

calculates values of n for which (c) is satisfied if and only if m is an odd integer.

What has NOT yet been proved is that no value of n exists which satisfies (c) other than those calculated from (d).

the solution of $48N^2 - E^2 = -1$ has the form
$$E=\frac{\left ( 7+\sqrt{48} \right )^k+\left ( 7-\sqrt{48} \right )^k}{2}$$

for k=0,1,2,... (not only the odd numbers)

so other number that not has the form of E=... , is not a solution.

then because E=4n+3 , we need to discard some values for k.

PD: a good book in Diophantine Equations is: "https://diendantoanhoc.net/index.php?app=core&module=attach&section=attach&attach_id=7203"
in page 119 talk about the function $Ax^2 - y^2 = -1$

 

[Buzz Bloom]

Hi @MAGNIBORO:

Thank you for posting the link to PDF file with the discussion of Diophantine equations. It looks like exactly what I was hoping to see to help get a deeper understanding of this kind of problem.

Regards,
Buzz