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 a2+b2 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 = 32 
24 and hence you can show 144 divides a2+b2 if you can show that 32 and 24 each divide a2+b2 . I'll do 32 and give you an idea on how to approach 24.
a2+b2 is divisible by 21 and hence it is divisible by 3. Thus
a2 + b2 = 3
for some integer k.
From the equation above, if a is divisible by 3 so is b2 and hence b is divisible by 3. Thus a2 and b2 are both divisible by 9 and therefore a2+b2 is divisible by 9. Likewise if b is divisible by 3 then a2 and b2 are both divisible by 9 and therefore a2+b2 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 a2 by 3 is also 1. If the remainder on dividing a by 3 is 2 then the remainder on dividing a2 by 3 is also 1. The same is true for b. If b is not divisible by 3 ten the remainder, on dividing b2 by 3 is 1. Finally if both a and b are not divisible by 3 then the remainder, on dividing a2+b2 by 3 is 1+1=2 and hence a2+b2 is not divisible by 3.
The result here is then that if a2+b2 is divisible by 3 it is also divisible by 32.
The remainders are the key to showing that if a2+b2 is divisible by 7 it is also divisible by 24 . What are the possible remainders when a is divided by 7? What are the possible remainders when a2 is divided by 7?
Penny