The centred residue quadruple of $pm=a^2+b^2+c^2+d^2$ has norm $mn$ with $1\le n<m$

Since $m>1$ the hypothesis $m\ge 1$ of [L1] holds, so $a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }},d^{\mathrm{\prime }}$ are defined and satisfy $a\equiv a^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}m)$, $b\equiv b^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}m)$, $c\equiv c^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}m)$, $d\equiv d^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}m)$ together with $-m<2a^{\mathrm{\prime }}\le m$, $4a^{\mathrm{\prime }2}\le m^2$ and the same three conditions for $b^{\mathrm{\prime }}$, $c^{\mathrm{\prime }}$ and $d^{\mathrm{\prime }}$.

By [L2] applied to the products $a\cdot a$, $b\cdot b$, $c\cdot c$, $d\cdot d$ and then to the sums, $a^{\mathrm{\prime }2}+b^{\mathrm{\prime }2}+c^{\mathrm{\prime }2}+d^{\mathrm{\prime }2}\equiv a^2+b^2+c^2+d^2=pm(\mathrm{m}\mathrm{o}\mathrm{d}m)$, and $pm\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}m)$ since $m\mid pm$; so by [F2] the modulus $m$ divides $a^{\mathrm{\prime }2}+b^{\mathrm{\prime }2}+c^{\mathrm{\prime }2}+d^{\mathrm{\prime }2}$, and by [F3] there is an integer $n$ with $a^{\mathrm{\prime }2}+b^{\mathrm{\prime }2}+c^{\mathrm{\prime }2}+d^{\mathrm{\prime }2}=mn$.

The left-hand side of step 2.1 is a sum of squares, hence at least $0$, and $m>0$, so $n\ge 0$.

Summing the four bounds $4a^{\mathrm{\prime }2}\le m^2$, $4b^{\mathrm{\prime }2}\le m^2$, $4c^{\mathrm{\prime }2}\le m^2$, $4d^{\mathrm{\prime }2}\le m^2$ of step 1.1 gives $4mn\le 4m^2$, hence $mn\le m^2$ and, dividing by the positive integer $m$, $n\le m$.

If $n=0$ then $a^{\mathrm{\prime }2}+b^{\mathrm{\prime }2}+c^{\mathrm{\prime }2}+d^{\mathrm{\prime }2}=0$ forces $a^{\mathrm{\prime }}=b^{\mathrm{\prime }}=c^{\mathrm{\prime }}=d^{\mathrm{\prime }}=0$, so [F2] and step 1.1 give $m\mid a$, $m\mid b$, $m\mid c$ and $m\mid d$; writing $a=m\alpha$, $b=m\beta$, $c=m\gamma$, $d=m\delta$ with [F3] then gives $pm=m^2({\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2)$.

If $n=m$ then step 3.2 holds with equality, so each of the four bounds of step 1.1 is an equality: $4a^{\mathrm{\prime }2}=m^2$ and likewise for $b^{\mathrm{\prime }}$, $c^{\mathrm{\prime }}$, $d^{\mathrm{\prime }}$.

In the case $n=0$, cancelling the nonzero factor $m$ in step 4.1 by [L4] gives $p=m({\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2)$, so $m$ is a positive divisor of $p$ and [F1] forces $m=1$ or $m=p$, both excluded by $1<m<p$; hence $n\ge 1$.

In the case of step 4.2, $m^2=(2a^{\mathrm{\prime }}{)}^2$ with $m>0$ gives $2a^{\mathrm{\prime }}=m$ or $2a^{\mathrm{\prime }}=-m$, and the normalisation $-m<2a^{\mathrm{\prime }}\le m$ of step 1.1 leaves $2a^{\mathrm{\prime }}=m$; so $m=2s$ with $s:=a^{\mathrm{\prime }}$ a positive integer, and the same argument gives $b^{\mathrm{\prime }}=c^{\mathrm{\prime }}=d^{\mathrm{\prime }}=s$.

Still in the case $n=m$, $a\equiv s(\mathrm{m}\mathrm{o}\mathrm{d}m)$ gives $a=s+mt=s(1+2t)$ for some integer $t$ by [F2] and [F3], so $a^2-s^2=s^2((1+2t{)}^2-1)=4s^{2}t(t+1)$, which $m^2=4s^2$ divides; the same holds for $b$, $c$ and $d$.

In the same case, summing the four differences of step 6.1 and using [L3], $m^2$ divides $(a^2+b^2+c^2+d^2)-4s^2=pm-m^2$, and $m^2$ divides $m^2$, so $m^2\mid pm$.

Still in the case $n=m$, writing $pm=m^{2}k$ as [F3] permits and cancelling the nonzero factor $m$ by [L4] gives $p=mk$, so $m$ is a positive divisor of $p$ and [F1] forces $m=1$ or $m=p$, both excluded by $1<m<p$; hence $n\mathrm{\ne }m$.

Therefore $n\ge 1$ by step 5.1, $n\le m$ by step 3.2 and $n\mathrm{\ne }m$ by step 8.1, that is $1\le n<m$, with $a^{\mathrm{\prime }2}+b^{\mathrm{\prime }2}+c^{\mathrm{\prime }2}+d^{\mathrm{\prime }2}=mn$ from step 2.1.