Can any number of squares sum to a square?

Question

Suppose

$$a^2 = \sum_{i=1}^k b_i^2$$

where $a, b_i \in \mathbb{Z}$, $a>0, b_i > 0$ (and $b_i$ are not necessarily distinct).

Can any positive integer be the value of $k$?

The reason I am interested in this: in a irreptile tiling where the smallest piece has area $A$, we have $a^2A = \sum_{i=1}^k b_i^2A$, where we have $k$ pieces scaled by $b_i$ to tile the big figure, which is scaled by $a$. I am wondering what constraints there are on the number of pieces.

Here is an example tiling that realizes $4^2 = 3^2 + 7 \cdot 1^2$, so $k = 8$.

Fair enough. Obviously I haven't had enough coffee yet this morning. — Michael Seifert, Dec 07 '19 at 14:00
It might be a little less confusing to phrase this request as the values of $k$ for which $a^2 = \sum_{i = 1}^k b_i^2$, where the various $b_i$ values may or may not be distinct. For example, $4^2 = 3^2 + 1^2 + 1^2+ 1^2+ 1^2+ 1^2+ 1^2+ 1^2.$ — Michael Seifert, Dec 07 '19 at 14:02
To make the problem nontrivial, we must assume that the $b_i$ are not zero.I think that the case $k_1=k_2=\cdots k_n=1$ is always solvable in positive integers which would already solve the problem. — Peter, Dec 07 '19 at 14:05
@MichaelSeifert @ Peter Thanks, I incorporated both your feedback. — Herman Tulleken, Dec 07 '19 at 14:08
Note that for any even $k \geq 4$, we have a solution of the form $(n+1)^2 = n^2 + (2n + 1) \cdot 1^2$; and any as you noted, any Pythagorean triple solves $k = 2$. The interesting cases are when $k$ is odd. — Michael Seifert, Dec 07 '19 at 14:15
To show that $k_1=k_2=\cdots k_n=1$ always works we can use the idendity $$(2n+1)^2+(2n(n+1))^2=(2n^2+2n+1)^2$$ which allows us to add a square to an arbitary odd square to get an odd square again. — Peter, Dec 07 '19 at 14:29
@Peter as I understand it that is the basis of J.G.'s answer. — Herman Tulleken, Dec 07 '19 at 14:31
@HermanTulleken Well possible, I just wanted to point it out. — Peter, Dec 07 '19 at 14:32
My brain parsed "Is any value of $k$ possible?" as "Is there any value of $k$?" — L. F., Dec 08 '19 at 10:52
I don't know if can help, but there is the well know espression: $1^3 + 2^3 + \dots + n^3 = (1+2+\dots+n)^2$ — vi pa, Dec 08 '19 at 23:18
Then any natural number can be espressed as the sum of 4 square. — vi pa, Dec 08 '19 at 23:52
There is the theorem of Legendre that say that if and only if the integer is not of the form $n=4^a(8b+7)$ for $a,b∈\mathbb{Z}$ then it can be espressed as the sum of 3 squares. See wiki. — vi pa, Dec 09 '19 at 09:06

score 35 · Answer 1 · answered Dec 07 '19 at 17:34

"Is any $k$ possible?" An easy route to "Yes": You know from the Pythagorean theorem that two squares can add to a perfect square. $$c^2=a^2+b^2$$

$c^2$ must be either odd or even. If odd, it is the difference between two consecutive squares. $$c^2=2n-1=n^2-(n-1)^2$$ If even, $c^2$ is divisible by $4$ and is also the difference between two squares. $$c^2=4n=(n+1)^2-(n-1)^2$$ So in either case, $c^2$ equals the difference between two squares. $$c^2=r^2-s^2 \\ r^2=c^2+s^2=a^2+b^2+s^2$$ Here, $r^2$ is the sum of three squares.

This can be repeated indefinitely, increasing by one the number of squares in the sum which adds to a square. There is no limit to the number of squares that can be accumulated in the sum.

Actually, since an odd square is always $1$ modulo $4$, the $n$ in $c^2=2n-1$ is also always odd, so starting from odd $c^2$ always gives odd squares (e.g. starting from $5^2=4^2+3^2$, we get $5^2 = 13^2-12^2$, $13^2 = 85^2-84^2$...). — JiK, Dec 08 '19 at 01:49
@JiK It's possible to start with a non-primitive Pythagorean triple such as $10^2=8^2+6^2$, so some provision for even squares must be considered. — Keith Backman, Dec 08 '19 at 04:11

J.G. · Answer 2 · 2023-08-14T14:36:07.700

Yes, $k$ can be arbitrary. Define a sequence$$a_1:=3,\,a_{k+1}:=\frac12\left(a_k^2+1\right)$$ of odd positive integers (since $\frac12((2n+1)^2+1)=2(n^2+n)+1$), so$$\begin{align}a_{k+1}^2-a_k^2&=\frac14\left[(a_k^2+1)^2-4a_k^2\right]\\&=\left[\frac12(a_k^2-1)\right]^2\end{align}$$is a perfect square. Now define$$b_1:=3,\,b_{k+1}:=\frac12(a_k^2-1)$$so $a_k^2=\sum_{i=1}^kb_i^2$ for all positive integers $k$. The sequence $a_n$ is called the Pythagorean spiral or OEIS A053630.

Bill Dubuque · Accepted Answer · 2020-11-25T11:33:50.870

This holds far more generally. OP is the special case $S$ = integer squares, which is closed under multiplication $\,a^2 b^2 = (ab)^2,\,$ and has an element that is a sum of $\,2\,$ others, e.g. $\,5^2 = 4^2+3^2 $.

Theorem $ $ If $\,S\,$ is a set of integers $\rm\color{#0a0}{closed}$ under multiplication then
$\qquad\qquad \begin{align}\phantom{|^{|^|}}\forall\,n\ge 2\!:\text{ there is a }\,t_n\in S\,&\ \text{that is a sum of $\,n\,$ elements of $\,S$ }\\[.1em] \iff\! \text{ there is a }\,t_2\in S\,&\ \text{that is a sum of $\,2\,$ elements of $\,S\!$}\\ \end{align}$

Proof $\ \ (\Rightarrow)\ $ Clear. $\ (\Leftarrow)\ $ We induct on $n$. The base case $\,n = 2\,$ is true by hypothesis, i.e. we are given that $\,\color{#c00}{a = b + c}\,$ for some $\,a,b,c\in S.\,$ If the statement is true for $\,n\,$ elements then

$$\begin{align} s_0 &\,=\, s_1\ \ +\ s_2\ + \cdots +s_n,\ \ \ \ \,{\rm all}\ \ s_i\,\in\, S\\ \Rightarrow\ s_0a &\,=\, s_1 a + s_2 a + \cdots +s_n\color{#c00} a,\ \ \color{#0a0}{\rm all}\ \ s_ia\in S \\[.1em] &\,=\, s_1 a + s_2 a + \cdots + s_n \color{#c00}b + s_n \color{#c00}c \end{align}\qquad\qquad$$

so $\,s_0 a\in S\,$ is a sum of $\,n\!+\!1\,$ elements of $S$, completing the induction.

Remark $ $ A comment asks for further examples. Let's consider some "minimal" examples. $S$ contains $\,a,b,c\,$ wth $\,a = b + c\:$ so - being closed under multiplication - $\,S\,$ contains all products $\,a^j b^j c^k\ne 1$. But these products are already closed under multiplication so we can take $S$ to be the set of all such products. Let's examine how the above inductive proof works in this set.

$$\begin{align} \color{#c00}a &= b + c\\ \smash{\overset{\times\ a}\Longrightarrow}\qquad\qquad\, a^2 = ab+\color{#c00}ac\, &= b(a+c)+c^2\ \ \ {\rm by\ substituting}\,\ \color{#c00}a = b+c\\ \smash{\overset{\times\ a}\Longrightarrow}\ \ a^3 = b(a^2+ac)+ \color{#c00}ac^2 &= b(a^2+ac+c^2)+c^3 \\[.4em] {a^{n}} &\ \smash{\overset{\vdots_{\phantom{|^|}\!\!}}= \color{#0a0}b (a^{n-1} + \cdots + c^{n-1}) + c^{n}\ \ \text{ [sum of $\,n\!+\!1\,$ terms]}}\\[.2em] {\rm by}\ \ \ a^{n}-c^{n} &= (\color{#0a0}{a\!-\!c}) (a^{n-1} + \cdots + c^{n-1})\ \ \ {\rm by}\ \ \color{#0a0}{b = a\!-\!c} \end{align}\quad\ \ \ $$

So the proof's inductive construction of an element that is a sum of $n+1$ terms boils down here to writing $\,a^n\,$ that way using the above well known factorization of $\,a^n-c^n\,$ via the Factor Theorem.

By specializing $\,a,b,c\,$ one obtains many examples, e.g. using $\,5^2,4^2,3^2$ as in the OP then $S$ is set of squares composed only of those factors, and the $\,n\!+\!1\,$ element sum constructed is

$$ 25^n =\, 9^n + 16(25^{n-1}+ \cdots + 9^{n-1})$$

This is a very clear generalization. Would be lovely if you could add one or two unexpected examples other than squares (I could only think of "multiple of m" and other powers which are not so interesting.) — Herman Tulleken, Dec 10 '19 at 00:28
@R.. Yes, it is indeed constructive, e.g. see the Remark I appended. — Bill Dubuque, Dec 11 '19 at 19:17

Herman Tulleken · Answer 4 · 2019-12-09T05:08:00.790

22

Here is a geometric solution (for $k > 5$).

The following are solutions for 6, 7, and 8 squares.

We can replace a square in each of these with four equal size squares to find a tiling with 3 more squares, so we can get 9, 10, and 11 squares. Repeating this we can get any number of squares larger than 5.

I show one iteration below:

edited Dec 09 '19 at 05:08

answered Dec 09 '19 at 04:56

Herman Tulleken

3,608

Beautiful and simple. Well done. – Eric Duminil Dec 09 '19 at 20:54
So the unit square is defined as whatever the smallest square is? – Cruncher Dec 09 '19 at 21:53
@Cruncher yes indeed. – Herman Tulleken Dec 09 '19 at 22:18
i.e. to add $,3,$ more squares, scale by $2^2$ then change one $(2a)^2$ to $a^2 + a^2 + a^2 + a^2\ \ $ – Bill Dubuque Dec 09 '19 at 22:59
Yes, although technically we don't need to scale since areas cancel out in the identity leaving only the integer squares. – Herman Tulleken Dec 09 '19 at 23:02
That's the algebraic essence of what is happening. It does require scaling (though that it hidden in the "proof by picture" handwaving). – Bill Dubuque Dec 09 '19 at 23:06
I was gonna disagree with you and as I wrote out the identity I saw the evil fraction that requires the scaling to get integers. – Herman Tulleken Dec 09 '19 at 23:11
3

Yes, when we split a square into 4 we are halving its side length, which would yield nonintegers if the side length was odd. To get around that we first scale everything by 2 to ensure all square side lengths (= summands) are even. Then it's precisely the algebraic transformation I mentioned above. – Bill Dubuque Dec 09 '19 at 23:16
1

I added an answer which highlights the key algebraic properties at the heart of the matter. – Bill Dubuque Dec 10 '19 at 00:14

score 3 · Answer 5 · answered Dec 09 '19 at 14:01

Yes.

For $k = 2$: $3^2 + 4^2 = 5^2$

For $k > 2$:
Start with a solution for $k-1$
Multiply both sides by $5^2$
Replace one $(5a)^2$ on the left with $(3a)^2$ + $(4a)^2$.

$3^2 + 4^2 = 5^2$
$15^2 + 20^2 = 25^2$
$9^2 + 12^2 + 20^2 = 25^2$ (k = 3)

$45^2 + 60^2 + 100^2 = 125^2$
$27^2 + 36^2 + 60^2 + 100^2 = 125^2$ (k = 4)

Repeat until k is as desired.

score 3 · Answer 6 · answered Dec 10 '19 at 19:00

It is known enough the generalization of Phytagorean triples which let us to say YES. Just as for $k=2$ two parameters are needed, for $k\gt2$ we need $k$ arbitrary parameters $t_1,t_2\cdots,t_k$ so we have the identity easily verified $$(t_k^2-t_1^2-t_2^2\cdots-t_{k-1}^2)^2+(2t_1t_k)^2+(2t_2t_k)^2+\cdots+(2t_{k-1}t_k)^2=(t_1^1+\cdots+t_k^2)^2$$.

(Note that for $k=2$ we have the quite known parameterization of Phytagorean triples).

David G. · Answer 7 · 2019-12-08T23:58:17.953

Solutions exist for every $k>0$. The simplest forms use most of the $b_n$ values as $1$. I will list them as "$b_*$". I suspect there are infinitely many distinct answers for each $k\ge2$, but I can't prove it.

if $k = 2n$ (k is even) and large enough ($k\ge4$ or $n>1$):

$ a = n $
$ b_1 = n-1 $
$ b_* = 1 $

\begin{align} \sum_{i=1}^k b_i^2 & = (n-1)^2 + (2n-1)\cdot1^1 \\ & = n^2 - 2n + 1 + 2n - 1 \\ & = n^2 \\ & = a^2 \\ \end{align}

if $k = 4n + 1$ and large enough ($k\ge9$ or $n>1$)

$ a = n+1 $
$ b_1 = n-1 $
$ b_* = 1 $

\begin{align} \sum_{i=1}^k b_i^2 & = (n-1)^2 + ((4n+1)-1)\cdot1^1 \\ & = n^2 - 2n + 1 + 4n \\ & = n^2 +2n + 1 \\ &= (n + 1)^2 \\ & = a^2 \\ \end{align}

if $k = 4n + 3$

$ a = 2n+3 $
$ b_1 = 2n+2 $
$ b_2 = 2 $
$ b_* = 1 $

\begin{align} \sum_{i=1}^k b_i^2 & = (2n+2)^2 + 2^2 + ((4n+3)-2)\cdot1^1 \\ & = 4n^2 + 8n + 4 + 4 + 4n + 1 \\ & = 4n^2 + 12n + 9 \\ &= (2n + 3)^2 \\ & = a^2 \\ \end{align}

The only values that don't work with these patterns are $k$ = 1, 2, or 5.

For $k=1$, $a=b_1$ for any values.

For $k=2$, we have we have a well known case, with minimal value $5^2=4^2+3^2$

For $k=5$, the minimal value is $4^2=3^2+2^2+1^2+1^2+1^2$

Maybe I am missing something. Sup $k = 9$, so $n = 2$, and $a = 3, b_1 = 2$, then we have $3^2 = 2^2 + 5\cdot 1$, but then $k = 1 + 5 = 6$ ... ? — Herman Tulleken, Dec 08 '19 at 18:23
@Herman If k=9, a=3, $b_1$=1, so we have $3^2 = 1^2 + 8 \cdot 1$ — David G., Dec 08 '19 at 20:22
You can maybe improve the answer by showing some calculations to save the reader from having to do them. — Herman Tulleken, Dec 08 '19 at 23:23

score 1 · Answer 8 · answered Jun 15 '21 at 16:23

Many of the other answers (based on extending smaller values of $k$) yield a fairly large value for the LHS. It’s worth pointing out that we actually know a great deal about the set of integers representable as the sum of $k$ nonzero squares.

See, for instance, this other answer, where it is shown that every integer $\ge 34$ is the sum of five positive squares. Immediately, by padding with $1$s, we can extend this to

For $k\ge 5$, every integer $> k+28$ can be written as the sum of $k$ nonnegative squares.

Therefore, we can just choose $a^2$ to be the first square greater than $k+28$, which makes $a^2 \le k + 2\sqrt k + O(1)$, which makes $$a \le \sqrt k + 1 + O(k^{-1/2}).$$ This basically optimal as clearly $a \ge \lceil \sqrt k\rceil^2$.

We might be able to shave down the constant $28$ somewhat by increasing the lower bound on $k$, but it can’t be reduced all the way to $0$ because of the gap between the first two nonzero squares $1$ and $4$. There must be some constant $C>0$ that is the optimal replacement for $28$ for all $k\ge k_0$ (I wouldn’t be surprised if it is already optimal).

score 0 · Answer 9 · answered Dec 07 '20 at 23:42

Yes, any number of squares can be a perfect square. Take the example of Pythagorean triples generated by the formula with a table of triples shown below it where $n$ is a set and where $k$ is a set member.

\begin{equation} A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2 \end{equation} \begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline \end{array} Since $3^2+4^2=5^2$, we can replace the $5$ in $5,12,13$ with $3,4$ so $3^3+4^2+12^2=13^2$. The value of $A$ can be any odd number greater than $1$ so there is always a triple where side-A of one equals side-C of another. Extending to a couple more squares, $3^2+4^2+12^2+84^2+132^2+12324=12325$. The process can be extended to infinite numbers of squares summed.

Let me know if you would like to know how to find "extensions".

Can any number of squares sum to a square?

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