Challenge Level

We received solutions to this problem from Andrei Lazanu from School No. 205 in Bucharest, Romania, from Fiona Watson and Sammy Lane from Stamford High School, from Chong Ching Tong, Chen Wei Jian and Teo Seow Tian from Class Secondary 1B in River Valley High School, in Singapore, and from Aftab Hussain. Well done to you all.

Everybody used the same arguments to arrive at their conclusions; here is what they said:

(2,4,6), (10,12,2) & (2,6,8) are all examples of triples of
whole numbers (a,b,c) such that a ^{2} + b ^{2} + c
^{2} is a multiple of 4:

2 ^{2} + 4 ^{2} + 6 ^{2} = 56

10 ^{2} + 12 ^{2} + 2 ^{2} = 248

2 ^{2} + 6 ^{2} + 8 ^{2} = 104

**To find out if a, b and c must all be even, let's first
consider what happens when all three numbers are
even:**

If any number is multiplied by 2, it is sure to be even.
So?

let | a = 2x | b = 2y | c = 2z |

then?

(2x) ^{2} + (2y) ^{2} + (2z) ^{2}
=

4x ^{2} + 4y ^{2} + 4z ^{2} =

4(x ^{2} + y ^{2} + z ^{2} )

which is a multiple of 4.

**Next, let's consider taking three odd
numbers:**

An odd number is 1 more than an even number. So?

let | a = 2x + 1 | b = 2y + 1 | c = 2z +1 |

then?

(2x + 1) ^{2} + (2y + 1) ^{2} + (2z + 1)
^{2} =

4x ^{2} + 4x + 1 + 4y ^{2} + 4y + 1+ 4z
^{2} + 4z + 1 =

4(x ^{2} + x + y ^{2} + y + z ^{2} + z) +
3

which is 3 more than a multiple of 4.

Therefore we cannot end up with a multiple of 4 if we start with
three odd numbers.

**Let's now consider taking two even numbers and one odd
number:**

let | a = 2x | b = 2y | c = 2z + 1 |

then?

(2x) ^{2} + (2y) ^{2} + (2z + 1) ^{2}
=

4x ^{2} + 4y ^{2} + 4z ^{2} + 4z + 1
=

4(x ^{2} + y ^{2} + z ^{2} + z) + 1

which is 1 more than a multiple of 4.

Therefore we cannot end up with a multiple of 4 if we start with
two even numbers and one odd number.

**Finally, let's consider taking two odd numbers and an
even number:**

let | a = 2x + 1 | b = 2y + 1 | c = 2z |

then?

(2x + 1) ^{2} + (2y + 1) ^{2} + (2z)
^{2} =

4x ^{2} + 4x + 1 + 4y ^{2} + 4y + 1 + 4z
^{2} =

4(x ^{2} + x + y ^{2} + y + z ^{2}) + 2

which is 2 more than a multiple of 4.

Therefore we cannot end up with a multiple of 4 if we start with
two odd numbers and one even number.

**Therefore it is necessary for all three numbers to be
even if the sum of their squares is to be a multiple of
4.**