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If a number can be expressed as sum of $2$ squares then every factor it can be expressed as sum of two squares

Mathematics Asked by Soham Chatterjee on November 15, 2020

Lemma : If a number which can be written as a sum of two squares is divisible by a number which is not a sum of two squares, then the quotient has a factor which is not a sum of two squares. (This is Euler’s second Proposition).

Let $a,b $ be two relatively prime numbers. Let $q $ is a proper factor of $a^2+b^2$ and $q $ can not be represented as sum of $2$ squares. Hence $q>2$.

Let $m,n $ be 2 numbers for which $mq,nq $ are the closest multiples of $q $ to $a,b $ respectively. Let $c=a-mq,d=b-nq $. Hence $|c|,|d|<frac{q}{2}$ and

$$a^2+b^2=(c+mq)^2+(d+nq)^2=m^2q^2+n^2q^2+2mqc+2nqc+c^2+d^2.$$ Hence $q$ divides $c^2+d^2$. Let $c^2+d^2=qr $. Let $g=gcd (c,d ) $. Now Let $m=gcd (g,q) $. Hence $m $ divides $q,c,d $. Hence $m $ divides $a,b $. As $a,b $ are relatively prime hence $m=1$.

Hence $g^2$ divides $r $. Let $e=frac{c}{g} ,f=frac{d}{g},s=frac{r}{g^2}$. Therefore $e^2+f^2=qs.$ Now, $$qs=e^2+f^2leq c^2+d^2 <frac{q^2}{4}+frac{q^2}{4}=frac{q^2}{2}.$$ Therefore $s <frac{q}{2}.$

Now finally, the descent step: if $q$ is not the sum of two squares, then by Lemma there must be a factor $q_{1}$ say of $s$ which is not the sum of two squares. But $q_{1} leq s<frac{q}{2} <q$ and so repeating these steps we shall be able to find a strictly decreasing infinite sequence $q, q_{1}, q_{2}, cdots$ of positive integers which are not
themselves the sums of two squares but which divide into a sum of two relatively prime squares. since such an infinite descent is impossible, we conclude that $q$ must be expressible as a sum of

Now my question is: How in the last paragraph

Now finally, the descent step: if $q$ is not the sum of two squares, then by Lemma there must be a factor $q_{1}$ say of $s$ which is not the sum of two squares. But $q_{1} leq s<frac{q}{2} <q$ and so repeating these steps we shall be able to find a strictly decreasing infinite sequence $q, q_{1}, q_{2}, cdots$ of positive integers which are not
themselves the sums of two squares but which divide into a sum of two relatively prime squares. since such an infinite descent is impossible, we conclude that $q$ must be expressible as a sum of

How we are getting an infinite sequesnce of numbers?

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