Pellbles

You find some pebbles and arrange them in groups of increasing diamond-shapes as follows:

You notice that the first and fourth groupings use a square number of pebbles ( 1 2 = 1 1^2 = 1 and 5 2 = 25 5^2 = 25 ).

If the pattern continues, how many other groupings would also use a square number of pebbles?

Infinitely more Finitely more (but more than zero) No more

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1 solution

Chris Lewis
Mar 10, 2021

The n th n^\text{th} diamond shape has D ( n ) = n 2 + ( n 1 ) 2 D(n)=n^2+(n-1)^2 dots, as illustrated below:

So the equation we want to solve is n 2 + ( n 1 ) 2 = m 2 n^2+(n-1)^2=m^2 where m , n m,n are natural numbers. Rearranging this, 2 n 2 2 n + 1 = m 2 4 n 2 4 n + 2 = 2 m 2 ( 2 n 1 ) 2 2 m 2 = 1 \begin{aligned} 2n^2-2n+1 &=m^2 \\ 4n^2-4n+2&=2m^2 \\ (2n-1)^2-2m^2&=-1 \end{aligned}

which is an example of Pell's equation (hence the problem's title). Since this equation has infinitely many solutions in integers, there are infinitely many diamonds made up of a square number of dots.

The next one is D ( 21 ) = 841 = 2 9 2 D(21)=841=29^2 .

Nice solution! Next challenge: can you come up with a recursive equation for it?

David Vreken - 3 months ago

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Let all the solutions of ( n , m ) (n,m) be denoted as ( n 1 , m 1 ) , ( n 2 , m 2 ) , ( n 3 , m 3 ) , (n_1, m_1), (n_2, m_2) , (n_3, m_3) , \ldots , where n 1 < n 2 < n_1 < n_2 < \cdots and m 1 < m 2 < m_1 < m_2 < \cdots .

Then for k 2 k \geqslant2 , we have the recursive relations

n k = 6 n k 1 n k 2 2 , n 1 = 1 , n 2 = 4 m k = 6 m k 1 m k 2 , m 1 = 1 , m 2 = 5 n_k = 6 n_{k-1} - n_{k-2} - 2, \quad n_1 = 1, n_2 = 4 \\ m_k = 6 m_{k-1} - m_{k-2}, \quad m_1 = 1, m_2 = 5

Voila: ( n , m ) = ( 1 , 1 ) , ( 4 , 5 ) , ( 21 , 29 ) , ( 120 , 169 ) , ( 697 , 985 ) , ( 4060 , 5741 ) , (n,m) = (1,1), (4,5), (21,29), (120,169), (697,985), (4060,5741), \ldots

Pi Han Goh - 3 months ago

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I combined some things to get n k + 1 = 3 n k + 2 2 n k 2 2 n k + 1 1 n_{k + 1} = 3n_k + 2\sqrt{2n_k^2 - 2n_k + 1} - 1 , which is the same sequence that you showed. By the way, this sequence is ( OEIS A046090 .

David Vreken - 3 months ago

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@David Vreken Now change the last two paragraph of the question to:

You notice that only the first grouping uses a cube number of pebbles ( 1 3 = 1 1^3 = 1 ).

Can we find any other such grouping?

The options:

[ ] Yes, finitely many
[ ] Yes, infinitely many
[x] No


I don't know how to prove this but I'm pretty confident of the answer.

Pi Han Goh - 3 months ago

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@Pi Han Goh I'm not sure how to prove it, either, but I did verify that your conjecture is true with a Python program for the first 10,000,000 groupings.

David Vreken - 3 months ago

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@David Vreken Nearly a week late to the party, but here goes: The equation we want to solve in integers is m 3 = 2 n 2 2 n + 1 m^3=2n^2-2n+1

Multiply by 8 8 : 8 m 3 = 16 n 2 16 n + 8 = ( 4 n 2 ) 2 + 4 8m^3=16n^2-16n+8=(4n-2)^2+4

Put x = 2 m x=2m and y = 4 n 2 y=4n-2 to get y 2 = x 3 4 y^2=x^3-4

which is an example of Mordell's equation . Better still, this precise example is solved on page 6 of this awesome paper by Keith Conrad. He shows the only solutions in integers to the above equation are ( x , y ) = ( 2 , ± 2 ) (x,y)=(2,\pm2) and ( x , y ) = ( 5 , ± 11 ) (x,y)=(5,\pm11) .

The first of these leads to solutions in the original equation ( m , n ) = ( 1 , 0 ) (m,n)=(1,0) and ( m , n ) = ( 1 , 1 ) (m,n)=(1,1) . The second gives non-integer values for ( m , n ) (m,n) so is discounted.


...and yes, I've spent way more of that week than I'd like to admit thinking about this question!

Chris Lewis - 2 months, 3 weeks ago

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@Chris Lewis Wow, nicely done!

David Vreken - 2 months, 3 weeks ago

@Chris Lewis You can't just submit your answer here. You have to submit a new problem and post your findings there!

Pi Han Goh - 2 months, 3 weeks ago

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@Pi Han Goh Haha, OK, it's posted now (though I really feel it was your question!).

This is obviously opening a can of worms, but what kind of intersections with other sequences of figurate numbers are possible?

Triangular numbers seem fairly easy (there's an obvious pattern if you solve 1 2 m ( m + 1 ) = 2 n 2 2 n + 1 \frac12 m(m+1)=2n^2-2n+1 for m m )

We've done squares, and cubes...how about square pyramidal numbers?

So, trying to solve 1 6 m ( m + 1 ) ( 2 m + 1 ) = 2 n 2 2 n + 1 \frac16 m(m+1)(2m+1)=2n^2-2n+1

A simple search shows there are some solutions...but how many?

Chris Lewis - 2 months, 3 weeks ago

Yup, but only through working some numbers out; the sequence defined by a 1 = 1 a_1=1 , a 2 = 4 a_2=4 and a i + 2 = 6 a i + 1 a i 2 a_{i+2}=6a_{i+1}-a_i-2

makes D ( a i ) D\left(a_i \right) a square number for all i i .

Chris Lewis - 3 months ago

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@Pi Han Goh - you just beat me to it! It's not too bad to prove that these always give solutions (induction works). Is there an easy way to show there are no other possible solutions?

Chris Lewis - 3 months ago

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@Chris Lewis What do you mean by no other possible solutions? Your solution clearly demonstrates that there are infinitely many solutions.

Pi Han Goh - 3 months ago

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@Pi Han Goh No other solutions than the ones generated by the recursive formulas. Like, how do we know there isn't a solution between ( 697 , 985 ) (697,985) and ( 4060 , 5741 ) (4060,5741) ? (Without just checking, obviously)

Chris Lewis - 3 months ago

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@Chris Lewis It's a Pell's equation. You can construct a general formula, (like Binet's formula), then convert it to a recursive formula.

Pi Han Goh - 3 months ago

An alternative approach is to note that n 1 , n , m n-1,n,m form a Pythagorean triple; this also leads to Pell's equation.

Chris Lewis - 3 months ago

lol I messed up I had the exact same equation except with (n+1) where m should have been. Agh.

chase marangu - 1 month, 4 weeks ago

Clever title! I get it now, Pell-bles haha

Mahdi Raza - 2 weeks, 5 days ago

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