[Math] The product of two numbers that can be written as the sum of two squares

number theory

Prove that if n is the product of two numbers that can be written as the sum of two squares then n can be written as the sum of two squares.

Best Answer

See the Brahmagupta–Fibonacci identity.

Or consider the formula for the product of two complex numbers, and take the moduli.

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Number Theory – Prove Every Odd Prime Number Can Be Written as a Difference of Two Squares

$1$. Let $(x+y)(x-y) = p$

Since $p$ is prime, the smaller divisor has to be one, ie. $(x-y) = 1$, giving $2y+1 = p \implies y = \frac{p-1}{2}$ (you're guaranteed y is an integer because $p$ is an odd number).

So the only possible solution set is $x = \frac{p+1}{2}, y = \frac{p-1}{2}$

$2$. Uniqueness already established via reasoning above.

$3$. Possible, but it will be non-unique as $(x-y)$ can take on multiple values, e.g. $1$ or a single prime divisor of $p$ or a product of some (but not all) prime divisors of $p$.

$4$. No, because again $(x-y)$ = 1 is forced. But now you get $x = \frac{3}{2}$ which is non-integral. So no integer solution sets exist.

[Math] Which rationals can be written as the sum of two rational squares

Let $r=\frac{p}{q}$, where $p$ and $q$ are integers. We will show that $r$ can be written as the sum of two rational squares if and only if $pq$ can be written as the sum of the squares of two integers.

Equivalently, if $p\ne 0$, then $\frac{p}{q}$ is a sum of the squares of two rationals if and only if every prime divisor of $pq$ of the form $4k+3$ occurs to an even power.

Proof: If $pq$ is the sum of the squares of two integers, it is clear that $r$ is the sum of the squares of two rationals.

For the other direction, suppose that $r$ can be written as the sum of the squares of two rationals. Without loss of generality we may assume that the rationals are $\frac{a}{c}$ and $\frac{b}{c}$ for some integers $a,b,c$. Then $$\frac{a^2+b^2}{c^2}=\frac{pq}{q^2}.$$ So $c^2pq$ is a sum of two squares. It follows that every prime of the form $4k+3$ occurs to an even degree in the prime power factorization of $c^2pq$, and hence of $pq$. It follows that $pq$ is a sum of two squares.

Best Answer

Related Solutions

Number Theory – Prove Every Odd Prime Number Can Be Written as a Difference of Two Squares

[Math] Which rationals can be written as the sum of two rational squares

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