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Well done to Andrew (Wilson's School), Shibon
(North London Collegiate School), and Jonathan (Oakwood Park
Grammar School), who had the right answer to the situation when the
squares are parallel to the lines on the grid.
Shibon concluded that:
It is impossible to have a square whose area is (4n + 3) as all
square numbers when divided by 4 have remainder of either 0 or 1 as
shown below. No square number divided by 4 is left with a remainder
of 3.
Andrew sent us a good proof:
Let n be the integer we wish to square.
If n is even... let n = 2m
\begin{eqnarray} n^2 &=& 4m^2 \end{eqnarray}
Therefore n is divisible by 4.
If n is odd... let n = 2m+1
\begin{eqnarray} n^2 &=& (2m+1)^2 &=& 4m^2 + 4m + 1
&=& 4(m^2 + m) + 1 \end{eqnarray}
which must therefore be one more than a multiple of 4.
Therefore any integer, when squared, will leave a remainder of
either 0 or 1 when divided by 4. Hence it is not possible to have a
remainder 3.
For the situation when squares make angles
with the lines of the grid, we have a good answer from Daniel
(Savile Park).
You can put right-angled triangles (sides a, b, and c) on any
square.
Using Pythagoras' theorem: a ² + b ² =
c ²
So area of square = a ² + b ²
We have:
even number ² = (2n) ² = 4n ²
(which is divisible by 4)
odd number ² = (2n + 1) ² = 4n ²
+ 4n + 1= 4(n ² + n) + 1 (leaving a remainder of 1 when
divided by 4)
Therefore:
even ² + even ² is divisible by 4
odd ² + odd ² leaves a remainder of 2 when
divided by 4
even ² + odd ² leaves a remainder 1 when
divided by 4
So a ² + b ² never has a remainder of 3
when divided by 4.
Hence the area of a square can't be 4n + 3