Show that for every prime $p\equiv 1\pmod 4$, there exists positive integers, $n,m$ such that

$$n(4m-p)-1\mid mp$$

Or equivalent, if we let $m= \frac{p+4k+3}{4}$,

$$n(4k+3)-1\mid p\left(\frac{p+4k+3}{4}\right)$$

Often we can find solutions to

$$n(4k+3)-1= p $$

but this isn’t always true. For example, there are no such solutions for $p=73$. Instead we can pick $n=3$, $k=1$, to get $20\mid 73\cdot 20$

Update: on further investigation this is simply a potential special case solution of the Erdos-Strauss conjecture since $l(m(4n-p)-1)=mp$ can be arranged to $\frac{4}{p} = \frac{1}{mnp}+\frac{1}{n}+\frac{1}{ln}$.

We solve the problem in the following way:

Proposition. Let $p\equiv 1\pmod 4$, then there exists positive integers $m,n$ such that $$ n(4m-p)-1 \mid m $$

Hence $n(4m-p)-1$ will also divide $mp$. This does not depend on $p$ being prime.


Proof. If $p=3$, we let $(m,n)=(1,2)$ so that $$ n(4m-p)-1 = 2(4-3)-1 = 1 $$

Otherwise, we may assume $p\equiv 1,2\pmod 3$. Since $p\equiv 1\pmod 4$, this means $p\equiv 1,5 \pmod{12}$.


Case 1: $p\equiv 1 \pmod {12}$

We split this into two cases instead: $p\equiv 1,13\pmod{24}$. First let $p=1+24r$. We set $(m,n)=(6r,1)$, so that $$ n(4m-p)-1 = (24r-(1+24r))-1 = -2 $$ which divides $m=6r$.

For the second case of $p=13+24r$, we set $(m,n)=(6r+4,1)$, giving $$ n(4m-p)-1 = (24r+16-(13+24r))-1 = 2 $$ which also divides $m=2(3r+2)$.


Case 2: $p \equiv 5 \pmod {12}$.

Similarly we split this into two cases, $p\equiv 5,17 \pmod{24}$. For $p=5+24r$, we set $(m,n) = (6r+2,1)$. This gives $$ n(4m-p)-1 = (24r+8-(5+24r))-1 = 2 $$ which divides $m=2(3r+1)$.

For the other case of $p=17+24r$, we use $(m,n) = (6r+4,1)$ to get $$ n(4m-p)-1 = (24r+16-(17+24r))-1=-2, $$ which also divides $m=2(3r+2)$.

$$ \tag*{$\square$} $$


Extras.

It seems trickier to require $n(4m-p)-1=dp$ for some positive integer $d$ instead. In particular, it is related to another unanswered problem. (The relationship being this is a sub-problem of that, which has a more general $p$ instead of prime.)

For the case of $p\equiv 5 \pmod 8$, there is a straightforward answer.


Case 1: $p\equiv 5 \pmod 8$

The approach is to set $n(4m-p)-1=p$.
Let $p\equiv 5 \pmod 8$ and write $p=5+8r$. Choose $n=2$ and set $$ m=\frac{1}{4}\left(\frac{p+1}{n}+p \right) $$ Then $$ \frac{p+1}{n} = \frac{6+8r}{2} = 3+4r\equiv 3 \pmod 4 $$ Since $p\equiv 1 \pmod 4$, $m$ is an integer.

Now we observe that $$ n(4m-p)-1 = n\left(\frac{p+1}{n}+p - p\right)-1= (p+1)-1=p $$ Therefore $$ n(4m-p)-1 = p \mid mp $$


Case 2: $p\equiv 1 \pmod 8$

The previous method cannot still work because $m$ is not integral for $p\equiv 1 \pmod 8$.


Connection to another problem

The other problem is as follows:

Let $ 5< p=a^2+b^2$ be odd and squarefree. Show that there exists positive integers $1\leq m$ and $7 \leq k\equiv 3\pmod 4$ such that $$ X^2-mkX+m\frac{(kp+1)}{4} = 0 $$ has integer roots.

Suppose we can always find a solution of the form $$ n(4m-p)-1 = dp \mid mp $$ for some positive integers $(m,n,d)$. Then $d$ divides $m$, so we may replace $m$ with $dm$. i.e. solving $$ n(4dm-p)-1 = dp \mid (dm)p $$ Let $n=k-d$ and $w=k-2d$, then this becomes $$ \begin{align*} n(4md-p)-1 &= dp\\ (k-d)(4md-p) &= dp+1\\ 4md(k-d) &= kp+1\\ m(2d)(2k-2d) &= kp+1\\ m(k-w)(k+w) &= kp+1\\ mk^2-kp-1 &= mw^2\\ m^2k^2-mkp-m &= (mw)^2 \end{align*} $$ Now since $m^2k^2-mkp-m$ is the discriminant $D$ of the equation $$ X^2-mkX+m\frac{kq+1}{4}=0 $$ This shows that the equation has solutions $$ \frac{mk\pm \sqrt{D}}{2} = \frac{mk \pm mw}{2} $$ which must be integral (either $m$ even or $k$ and $w$ can be shown to be odd).

We also require $k\equiv 3\pmod 4$, which can be shown as follows: $$ \begin{align*} n(4m-p)-1 &= dp\\ (k-d)(4m-p)-1 &= dp\\ (k-d)(-1)-1 &\equiv d \pmod 4\\ d-k-1 &\equiv d \pmod 4\\ k &\equiv 3 \pmod 4 \end{align*} $$ Since $k=n+d$ it is always positive, but there is a small problem that it might be equal to $3$.

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