is it true that is m and n are sums of two squares and m divides n, then n/m is a sum of two
squares? Prove or give a counterexample.
Solution
If \"the sum of two squares\" mean \"sum of two integers square,\" the answer is
obviously no. If n = a^2 + b^2, you can make m = (2a)^2 + (2b)^2 = 4n. m obviously divides n.
But n/m = 1/4, which is not a sum of any two integers square. If the two numbers are not
necessarily integers, the answer is yes. Since any value n/m can be written as (square root of
n/m)^2 + 0^2..
is it true that is m and n are sums of two squares and m divides n, .pdf
1. is it true that is m and n are sums of two squares and m divides n, then n/m is a sum of two
squares? Prove or give a counterexample.
Solution
If "the sum of two squares" mean "sum of two integers square," the answer is
obviously no. If n = a^2 + b^2, you can make m = (2a)^2 + (2b)^2 = 4n. m obviously divides n.
But n/m = 1/4, which is not a sum of any two integers square. If the two numbers are not
necessarily integers, the answer is yes. Since any value n/m can be written as (square root of
n/m)^2 + 0^2.
