Well done Richard Mycroft, age 16 from Melbourn Village College, Cambridgeshire and Selvan Gnanakumaran and Tony Cardell for your solutions.

We assume here that a and b are integers and a ²+b ² is divisible by 3.

Here is Richard's which uses modulus arithmetic.

Any number is either -1, 0, or 1 mod 3. For any number n we have

${\displaystyle n^2=0\hspace{1em}\mathrm{or}\hspace{1em}1mod3.}$

${\displaystyle \mathrm{As}-1^2=1,0^2=0\mathrm{and}{1}^2=1}$

Therefore n ² = 0 mod 3 if n = 0 mod 3, or n ²= 1 mod 3 if n = 1 or -1 mod 3.

If a ² + b ² is divisible by 3 then a ² + b ² = 0 mod 3. Now, N mod 3 + M mod 3 =N +M mod 3. Therefore if both a and b are not divisible by 3 then

${\displaystyle a^2+b^2=2=-1mod3.}$

If one of a and b is not divisible by 3 (but the other is) then:

In both cases a ² + b ² is not divisible by 3. So if a ² + b ² is divisible by 3 then both a and b are divisible by 3.

Here is Tony's proof which is of course equivalent. Any number n falls into one of three categories, having a remainder of 0, 1 or 2 when divided by 3. Therefore any number n ² will have a remainder 0 ² = 0, 1 ² = 1, or 2 ² = 4, also giving a remainder 1 when divided by 3. For a ²+b ² to be divisible by 3, it must be congruent to 0 mod 3. The only way for this to happen, since a ² and b ² can only be zero or one mod 3 and so their maximum sum is two, is for them both to be zero mod 3. If a ² and b ² are both zero mod 3, they are both divisible by three, and thus both a and b are divisible by three.

Selvan proved this result in a different way: First say that if a ² +b ² is a multiple of 3 then a ² + b ² = 3n , where n is a positive integer.

Now if b is not a multiple of 3, b can be expressed in the form 3x \pm 1, where x is an integer. Therefore $a^2+(3x\pm 1){}^2=3n$ so $a^2=3n-9x^2\pm 6x-1$ and therefore a ² is not a multiple of three as it is expressed as a multiple of three minus 1. Hence a is not a multiple of 3, and therefore a can be expressed in the form $3y\pm 1$ where y is a positive integer.

Therefore $(3y\pm 1){}^2=3n-9x^2\pm 6x-1$ and so

${\displaystyle 9y^2\pm 6y+1=3n-9x^2\pm 6x-1}$

${\displaystyle 9y^2\pm 6y+2=3n-9x^2\pm 6x}$

The right hand side of the expression is a multiple of three, but the left hand side however is clearly not, as it is a multiple of three add 2. Here is a contradiction.

When b was said to be a non multiple of three, it led to saying that a must also be a non multiple, and a contradiction occurred. Therefore neither a nor b can be non multiples of three, hence if a ² + b ² is a multiple of three then a and b are both multiples of three.