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Adithya from Lexden Primary School noted
that there is a pattern to the gap between square numbers, with the
gap increasing by two each time. Adithya also noticed a pattern in
column A; this was also spotted by James from Wilson's School,
Niteesh from Vidyashilp Academy, Dan and Linden from Wilson's
School, and James from Eltham College. Here is a picture of
Linden's spreadsheet showing the pattern:
Krystof from Uhelny Trh, Prague, made
some comments about which squares appear in each
Column A contains squares of odd numbers. Column D contains squares
of even numbers which are not divisible by four. Column H contains
squares of multiples of four.
An anonymous solver sent us these
Firstly I noticed that all of the squares of odd numbers were in
the left-hand column, column A. All of the squares of even
multiples of 4 were in the right-hand column, column H, while all
of the squares of other even numbers are in column D.
I can explain these by saying that the squares of multiples of $4$
are always multiples of $8$. The squares of any other even numbers
are always multiples of $4$, while the squares of odd numbers do
not all have a common factor and therfore are only divisble by
I noticed that the final digit of the odd squares was always $1$,
$5$ or $9$ and for the even squares always $4$, $6$ or $0$. I also
noticed that the first square number was in column A, then the
second in column D, the third back in column A and the fourth in
column H. I also noticed that in column A, which has the squares of
odd numbers, the rows between the squares increases by $1$ every
time, for example there are $0$ rows between $1$ and $9$, $1$ row
between $9$ and $25$, $2$ rows between $25$ and $49$ etcetera. This
therefore tells me that the difference between odd numbers is also
always a multiple of $8$.
I can also see that when each odd square is represented in the form
$2n+1$ $n$ is always a multiple of $4$ ($2\times4+1=9,
2\times40+1=81$ etc.) I can use examples to demonstrate this, for
example $13\times13=169$ which not only ends in $9$ but is
also $2\times84+1$, and $84$ is a multiple of $4$. Not only this,
the difference between $169$ and the previous odd square is $48$,
which is a multiple of $8$, just like the pattern suggested. The
patterns also work for the next square numbers, i.e.
$14\times14=196$ and $15\times15=225$.
Lots of fantastic ideas - can it be proved
that these patterns will continue?
Herschel from the European School of
Varese, and Simon who didn't give his school both sent us algebraic
explanations for some of the patterns. Here is Herschel's
The square numbers show a definite pattern:
$1^2=1$ is in column A.
$2^2=4$ is in column D.
$3^2=9$ is in column A.
$4^2=16$ is in column H.
This pattern (A,D,A,H) repeats itself throughout the series, albeit
with larger gaps (blank rows) as the numbers increase.
This can be proven algebraically:
Suppose we call $n$ any multiple of $4$ We can write any whole
number as either $n$, $n+1$, $n+2$ or $n+3$. Since $n$ is a
multiple of $4$, we can re-write $n$ as $4a$, so that all whole
numbers are either $4a$, $4a+1$, $4a+2$ or $4a+3$.
$(4a)^2 = 16a^2$, which is $8\times(2a^2)$, so it is a multiple of
$8$ and hence appears in column H.
$(4a+1)^2 = 16a^2+8a+1$ which is $8\times(2a^2+a) + 1$, so it is
$1$ more than a multiple of $8$ and hence appears in column
$(4a+2)^2 = 16a^2+16a+4$ which is $8\times(2a^2+2a) + 4$, so it is
$4$ more than a multiple of $8$ and hence appears in column
$(4a+3)^2 = 16a^2+24a+9$ which is $8\times(2a^2+3a+1) + 1$, so it
is $1$ more than a multiple of $8$ and hence appears in column
After $4a+3$, the next number is $4a+4$ which is $4(a+1)$, so we
increase $a$ by $1$ and continue the trend. This means the pattern
of highlighted squares is H,A,D,A (or A,D,A,H since we ignore
$0^2=0$) and this continues forever.
Well done to all those who submitted