Quandaries
and Queries 

Hi My name is Ampa and I´m a secondary student having trouble solving the following exercise: given natural numbers a and b such that a^{2}+b^{2} is divisible by 21, prove that the same sum of squares is also divisible by 441. Please help me to solve the exercise. thanks 

Hi Ampa, The prime factorization of 144 is 144 = 3^{2} 2^{4} and hence you can show 144 divides a^{2}+b^{2} if you can show that 3^{2} and 2^{4} each divide a^{2}+b^{2} . I'll do 3^{2} and give you an idea on how to approach 2^{4}. a^{2}+b^{2} is divisible by 21 and hence it is divisible by 3. Thus
for some integer k. From the equation above, if a is divisible by 3 so is b^{2} and hence b is divisible by 3. Thus a^{2} and b^{2} are both divisible by 9 and therefore a^{2}+b^{2} is divisible by 9. Likewise if b is divisible by 3 then a^{2} and b^{2} are both divisible by 9 and therefore a^{2}+b^{2} is divisible by 9. If a is not divisible by 3, then the remainder when dividing a by 3 is either 1 or 2. If the remainder on dividing a by 3 is 1 then the remainder on dividing a^{2} by 3 is also 1. If the remainder on dividing a by 3 is 2 then the remainder on dividing a^{2} by 3 is also 1. The same is true for b. If b is not divisible by 3 ten the remainder, on dividing b^{2} by 3 is 1. Finally if both a and b are not divisible by 3 then the remainder, on dividing a^{2}+b^{2} by 3 is 1+1=2 and hence a^{2}+b^{2} is not divisible by 3. The result here is then that if a^{2}+b^{2} is divisible by 3 it is also divisible by 3^{2}. The remainders are the key to showing that if a^{2}+b^{2} is divisible by 7 it is also divisible by 2^{4} . What are the possible remainders when a is divided by 7? What are the possible remainders when a^{2} is divided by 7? Penny 

