# Thread: Sums of squares exercise

1. ## Sums of squares exercise

Let $\displaystyle m_0$ be the least possible $\displaystyle m$ for which there are $\displaystyle x$ and $\displaystyle y$ for which

$\displaystyle x^2 + y^2 = mp$, where p is a prime and $\displaystyle p \equiv 1 (mod4)$. Also $\displaystyle p \nmid x$, $\displaystyle p \nmid y$, $\displaystyle 0<m<p$ .

That there are such x,y and m is the subject of the previous exercise.

Thus, $\displaystyle x_0^2+y_0^2 = m_0p$ (1)

Show $\displaystyle m_0 \nmid x_0$ and $\displaystyle m_0 \nmid y_0$ unless $\displaystyle m_0 = 1$.

By taking $\displaystyle m_0$ as a factor of both $\displaystyle x_0$ and $\displaystyle y_0$, I find, from (1) above, that $\displaystyle p$ must be factored by $\displaystyle m_0$. But p is prime. Therefore $\displaystyle m_0=1$.

I feel I must have overlooked something, because I have not used the hypothesis that $\displaystyle m_0$ is minimal. What have I overlooked, please?

The above problem is from GE Andrews's "Number Theory", ex. 2, Chapter 11-2.

2. Originally Posted by Nivla
Let $\displaystyle m_0$ be the least possible $\displaystyle m$ for which there are $\displaystyle x$ and $\displaystyle y$ for which

$\displaystyle x^2 + y^2 = mp$, where p is a prime and $\displaystyle p \equiv 1 (mod4)$. Also $\displaystyle p \nmid x$, $\displaystyle p \nmid y$, $\displaystyle 0<m<p$ .

That there are such x,y and m is the subject of the previous exercise.

Thus, $\displaystyle x_0^2+y_0^2 = m_0p$ (1)

Show $\displaystyle m_0 \nmid x_0$ and $\displaystyle m_0 \nmid y_0$ unless $\displaystyle m_0 = 1$.

By taking $\displaystyle m_0$ as a factor of both $\displaystyle x_0$ and $\displaystyle y_0$, I find, from (1) above, that $\displaystyle p$ must be factored by $\displaystyle m_0$. But p is prime. Therefore $\displaystyle m_0=1$.

I feel I must have overlooked something, because I have not used the hypothesis that $\displaystyle m_0$ is minimal. What have I overlooked, please?

The above problem is from GE Andrews's "Number Theory", ex. 2, Chapter 11-2.
without loss of generality suppose that $\displaystyle m_0 \mid x_0.$ if $\displaystyle m_0 > 1,$ then $\displaystyle \gcd(m_0,y_0)=d > 1$ because $\displaystyle m_0 \mid m_0p - x_0^2=y_0^2.$ now let $\displaystyle x_0=m_0x_1, \ m_0=dm_1, \ y_0=dy_1.$

then $\displaystyle x_0^2+y_0^2=m_0p$ will give us $\displaystyle d(m_1^2x_1^2 + y_1^2)=m_1p.$ thus $\displaystyle d \mid m_1.$ let $\displaystyle m_1=dm_2.$ then $\displaystyle (m_1x_1)^2 + y_1^2=m_2p,$ which contradicts minimality of $\displaystyle m_0$ because $\displaystyle m_0=d^2m_2 > m_2.$