Integer as Sums and Differences of Consecutive Squares

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Theorem

Every integer $k$ can be represented in infinitely many ways in the form:

$k = \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm m^2$

for some (strictly) positive integer $m$ and some choice of signs $+$ or $-$.


Proof

First we notice that:

\(\ds \) \(\) \(\ds \paren {m + 1}^2 - \paren {m + 2}^2 - \paren {m + 3}^2 + \paren {m + 4}^2\)
\(\ds \) \(=\) \(\ds \paren {-1} \paren {m + 1 + m + 2} + \paren {m + 3 + m + 4}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds 4\)


Now we prove a weaker form of the theorem by induction:

Every positive integer $k$ can be represented in at least one way in the form:
$k = \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm m^2$
for some (strictly) positive integer $m$ and some choice of signs $+$ or $-$.

Basis for the Induction

From the identity above we have:

\(\ds 0\) \(=\) \(\ds 4 - 4\)
\(\ds \) \(=\) \(\ds + 1^2 - 2^2 - 3^2 + 4^2 - 5^2 + 6^2 + 7^2 - 8^2\)

Also we have, more or less trivially:

$1 = + 1^2$
$2 = - 1^2 - 2^2 - 3^2 + 4^2$
$3 = - 1^2 + 2^2$

This is the basis for the induction.


Induction Hypothesis

This is the induction hypothesis:

For some positive integer $k$, we have:
$k = \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm m^2$
for some (strictly) positive integer $m$ and some choice of signs $+$ or $-$.

Now we need to show true for $n = k + 4$:

$k + 4 = \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm {m_2}^2$
for some (strictly) positive integer $m_2$ and some choice of signs $+$ or $-$.


Induction Step

This is the induction step:

\(\ds \) \(\) \(\ds k + 4\)
\(\ds \) \(=\) \(\ds \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm m^2 + 4\) from induction hypothesis
\(\ds \) \(=\) \(\ds \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm m^2 + \paren {m + 1}^2 - \paren {m + 2}^2 - \paren {m + 3}^2 + \paren {m + 4}^2\) from the above identity

which is of the required form.

Hence every positive integer $k$ can be represented in at least one way in the required form.

$\Box$


From the identity above we can also conclude:

\(\ds 0\) \(=\) \(\ds 4 - 4\)
\(\ds \) \(=\) \(\ds \paren {m + 1}^2 - \paren {m + 2}^2 - \paren {m + 3}^2 + \paren {m + 4}^2 - \paren {m + 5}^2 + \paren {m + 6}^2 + \paren {m + 7}^2 - \paren {m + 8}^2\)

therefore if $k$ can be written as:

$k = \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm m^2$

it can also be written as:

$k = \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm m^2 + \paren {m + 1}^2 - \paren {m + 2}^2 - \paren {m + 3}^2 + \paren {m + 4}^2 - \paren {m + 5}^2 + \paren {m + 6}^2 + \paren {m + 7}^2 - \paren {m + 8}^2$

and:

$k = \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm \paren {m + 8}^2 + \paren {m + 9}^2 - \paren {m + 10}^2 - \paren {m + 11}^2 + \paren {m + 12}^2 - \paren {m + 13}^2 + \paren {m + 14}^2 + \paren {m + 15}^2 - \paren {m + 16}^2$

and so on.

Hence every positive integer $k$ can be represented in infinitely many ways in the required form.

$\Box$


Finally we cover the case where $k$ is a negative integer.

Since $-k$ is a positive integer, it can be represented in infinitely many ways in the required form.

For every representation of $-k$ in this way, by flipping every $+$ and $-$ sign, we end up with a representation of $- \paren {- k} = k$.

Hence every integer $k$, positive or negative, can be represented in infinitely many ways in the required form.

$\blacksquare$


Example

Since the proof above is constructive, we can follow the proof and derive:

\(\ds 15\) \(=\) \(\ds 3 + 4 + 4 + 4\)
\(\ds \) \(=\) \(\ds \paren {- 1^2 + 2^2} + \paren {3^2 - 4^2 - 5^2 + 6^2} + \paren {7^2 - 8^2 - 9^2 + 10^2} + \paren {11^2 - 12^2 - 13^2 + 14^2}\)

and its associated family of solutions:

\(\ds 15\) \(=\) \(\ds - 1^2 + 2^2 + 3^2 - 4^2 - 5^2 + 6^2 + 7^2 - 8^2 - 9^2 + 10^2 + 11^2 - 12^2 - 13^2 + 14^2 + 15^2 - 16^2 - 17^2 + 18^2 - 19^2 + 20^2 + 21^2 - 22^2\)

and so on.

However this may not give the least number of squares that this works for.

In fact we have:

\(\ds 15\) \(=\) \(\ds 1^2 - 2^2 + 3^2 - 4^2 + 5^2\)

from Triangular Number as Alternating Sum and Difference of Squares.


Also see


Historical Note

Wacław Sierpiński attributes this result to Paul Erdős and János Surányi.


Sources

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