Grid Lockout

Age 14 to 16

Challenge Level Yellow star

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

Number and algebra

Geometry and measure

Probability and statistics

Working mathematically

Advanced mathematics

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Grid Lockout

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