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

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

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

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

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 $20mid 73cdot 20$

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 $pequiv 1pmod 4$, 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 $pequiv 1,2pmod 3$. Since $pequiv 1pmod 4$, this means
$pequiv 1,5 pmod{12}$.

Case 1: $pequiv 1 pmod {12}$

We split this into two cases instead: $pequiv 1,13pmod{24}$. 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, $pequiv 5,17 pmod{24}$. 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 $pequiv 5 pmod 8$, there is a straightforward answer.

Case 1: $pequiv 5 pmod 8$

The approach is to set $n(4m-p)-1=p$.

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

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

Case 2: $pequiv 1 pmod 8$

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

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 $1leq m$ and $7 leq kequiv 3pmod 4$ such that
$$
X^2-mkX+mfrac{(kp+1)}{4} = 0
$$
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
$$
X^2-mkX+mfrac{kq+1}{4}=0
$$
This shows that the equation has solutions
$$
frac{mkpm sqrt{D}}{2} = frac{mk pm mw}{2}
$$
which must be integral (either $m$ even or $k$ and $w$ can be shown to be odd).

We also require $kequiv 3pmod 4$, 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 $3$.