Positive integral solutions to $\frac{m(m+5)}{n(n+5)}=p^2$, where $p$ is prime.

Question

Find all $(m,n)$ positive integers such that $\frac {m(m+5)}{n(n+5)}= p^2$ such that $p$ is a prime number.

I tried a lot but could not find a way to start, i cannot proceed with mod bashing please help!

Answer 1

This does not answers your question in full, it is just a start of the analysis of the problem and it can be a way to start.

From $\dfrac{m(m+5)}{n(n+5)}=p^2$ we obtain $m(m+5)=p^2n(n+5)$.

If you view this as a quadratic equation in $n$ we obtain:

$p^2n^2+5p^2n-m(m+5)=0$.

The solutions are $n=\dfrac{-5p^2+\sqrt{25p^4+4p^2m(m+5)}}{2p^2}$

The expression under the square-root sign must be a square of an integer so we have: $25p^4+4p^2m(m+5)=w^2$

If you view even this as a quadratic equation in the variable $p^2=r$ you obtain:

$25r^2+4rm(m+5)-w^2=0$

The solutions are $r=\dfrac{-4m(m+5)+\sqrt{(4m(m+5))^2+(10w)^2}}{50}$

The expression under the square-root sign must be a square of an integer so we have:

$(4m(m+5))^2+(10w)^2=h^2$

so, we obtained a Pythagorean triples and so much is known about them.

Now you can find, for example, on Wikipedia, what form must Pythagorean triples have and deduce something about the solutions.

There are many conditions here and from all of them you can say something about the solutions.

Answer 2

Numerically:

? for(m=2,10^5,for(n=1,m-1,p2=m*(m+5)/(n*(n+5));if(issquare(p2)&&p2==floor(p2),p=sqrtint(p2);if(isprime(p),print1("("n","m","p"), ")))))
(1,3,2), (2,9,3), (1,10,5), (7,16,2), (6,22,3), (9,121,11), (19,147,7),

Let $n$ is parameter, $2m+5\to x$, $2p\to y$, then solving of Pell equation $x^2-n(n+5)y^2=25$ can help us filter out solutions of source problem.

Answer 3

Below is a proof that the following is the full list of solutions: $$\begin{array}{rrr} p&m&n\\ \hline 2&3&1\\ 2&16&7\\ 3&9&2\\ 3&22&6\\ 5&10&1\\ 7&147&19\\ 11&121&9\\ \end{array}$$

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Let $m$ and $n$ be positive integers and $p$ a prime number such that $$\frac{m(m+5)}{n(n+5)}=p^2.$$ Setting $a:=2m+5$ and $b:=2n+5$ we find that $$p^2=\frac{m(m+5)}{n(n+5)}=\frac{\tfrac{a-5}{2}\tfrac{a+5}{2}}{\tfrac{b-5}{2}\tfrac{b+5}{2}}=\frac{a^2-25}{b^2-25},$$ and in particular that $p^2$ divides $(a-5)(a+5)$.

Case 0: $p^2$ divides neither factor.

In this case $p$ divides both factors then it divides $\gcd(a-5,a+5)=\gcd(a,10)$, so $p=2$ or $p=5$.

Subcase 0.1: If $p=2$ then we find that $$b^2=\frac{a^2-25}{p^2}+25=m(m+5)+25=m^2+5m+25,$$ which shows that $(m+2)^2<b^2<(m+5)^2$. This yields the two equations $$m^2+5m+25=b^2=(m+3)^2=m^2+6m+9,$$ $$m^2+5m+25=b^2=(m+4)^2=m^2+8m+16.$$ These have only the solutions $m=16$ and $m=3$, respectively, with $n=7$ and $n=1$ respectively. Hence in this case we find only the solutions $$(m,n,p)=(3,1,2)\qquad\text{ and }\qquad(m,n,p)=(16,7,2).$$

Subcase 0.2: If $p=5$ then we find that $5\mid a$ and so $a=5c$, yielding $$b^2=\frac{a^2-25}{p^2}+25=c^2+24.$$ The only solutions are $(b,c)=(5,1)$ and $(b,c)=(7,5)$. Only the latter corresponds to a solution to the original problem, which is $$(m,n,p)=(10,1,5).$$

These three solutions are the only solutions for which $p$ divides both factors. If $p$ does not divide both factors, then one of the factors is divisible by $p^2$.

Case 1: $p^2$ divides $a-5$

In this case $a=kp^2+5$ for some integer $k$. As $m$ is positive we have $a\geq7$ which shows that $k>0$. Then $$b^2=\frac{a^2-25}{p^2}+25=k^2p^2+10k+25>k^2p^2.$$ But the next square after $k^2p^2$ is $$(kp+1)^2=k^2p^2+2kp+1,$$ which shows that $10k+25\geq2kp+1$.

Subcase 1.0: If $p>10$ then there are only finitely many values of $k$ that satisfy the inequality above. In fact for $p>17$ we see that $$2kp+1>34k+1\geq10k+25,$$ because $k$ is positive. Hence $p\leq17$ and we are left with only finitely many cases to check; a quick check yields only the solutions $$(a,b,p)=(294,18,17),\qquad\text{ and }\qquad (a,b,p)=(247,23,11).$$ Only the latter corresponds to a solution to the original problem; it yields $$(m,n,p)=(121,9,11).$$

Subcase 1.1: If $p=7$ then we find that $$b^2=49k^2+10k+25=(7k+3)^2-32k+16<(7k+3)^2,$$ so either $b=7k+1$ or $b=7k+2$, yielding the two equations $$49k^2+10k+25=(7k+1)^2=49k^2+14k+1,$$ $$49k^2+10k+25=(7k+2)^2=49k^2+28k+4.$$ The latter yields no solutions, the former yields only $k=6$ corresponding to the triplet $(a,b,p)=(299,43,7)$, which in turn corresponds to the solution $$(m,n,p)=(147,19,7).$$

Subcase 1.2: If $p=5$ then we find that $$b^2=25k^2+10k+25=5(5k^2+2k+5),$$ which shows that $5\mid b$ and hence in turn $5\mid k$. Let $b=5c$ and $k=5l$ so that $$c^2=25l^2+2l+1.$$ Then $(5l)^2<c^2<(5l+1)^2$, a contradiction. So this yields no new solutions.

Subcase 1.3: If $p=3$ then we find that $$b^2=9k^2+10k+25,$$ which shows that $(3k+1)^2<b^2<(3k+4)^2$, yielding the two equations $$9k^2+10k+25=b^2=(3k+2)^2=9k^2+12k+4,$$ $$9k^2+10k+25=b^2=(3k+3)^2=9k^2+18k+9.$$ The former has no solutions, the latter has only the solution $k=2$ corresponding to $(a,b,p)=(23,9,3)$. This in turn corresponds to the following solution to the original problem: $$(m,n,p)=(9,2,3).$$

Subcase 1.4: If $p=2$ then we find that $$b^2=4k^2+10k+25,$$ which shows that $(2k+2)^2<b^2<(2k+4)^2$, yielding the equation $$4k^2+10k+25=b^2=(2k+3)^2=4k^2+12k+9.$$ It has the unique solution $k=8$ corresponding to $(a,b,p)=(37,19,2)$, in turn yielding $$(m,n,p)=(16,7,2),$$ which we already found.

Case 2: $p^2$ divides $a+5$

In this case $a=kp^2-5$ for some integer $k$. Then $k$ is positive because $a$ is, and we have $$b^2=\frac{a^2-25}{p^2}+25=k^2p^2-10k+25.$$ In particular $b<kp$ if $-10k+25<0$, i.e. if $k>2$. If $k\leq2$ then either $$b^2=p^2+15\qquad\text{ or }\qquad b^2=4p^2+5.$$ The latter has no solutions, the former has the unique solution $(p,b)=(7,8)$ corresponding to the solution $(a,b,p)=(44,8,7)$, which does not correspond to a solution to the original problem.

If $k>2$ then $b\leq kp-1$ because $$b^2=k^2p^2-10k+25<(kp)^2,$$ and hence $2(p-5)k\leq-24$ because $$k^2p^2-10k+25\leq(kp-1)^2=k^2p^2-2kp+1.$$ This means that $p<5$, i.e. $p=2$ or $p=3$.

Subcase 2.1: If $p=3$ then we find that $$b^2=9k^2-10k+25,$$ which shows that $(3k-2)^2<b^2<(3k+1)^2$, yielding the equations $$9k^2-10k+25=b^2=(3k-1)^2=9k^2-6k+1,$$ $$9k^2-10k+25=b^2=(3k)^2=9k^2.$$ The former has the unique solution $k=6$, the latter has no solutions. This yields the solution $(a,b,p)=(49,17,3)$, corresponding to $$(m,n,p)=(22,6,3).$$

Subcase 2.2: If $p=2$ then we find that $$b^2=4k^2-10k+25,$$ which shows that $b$ is odd and $(2k-3)^2<b^2<(2k+3)^2$, yielding the equations $$4k^2-10k+25=b^2=(2k-1)^2=4k^2-4k+1,$$ $$4k^2-10k+25=b^2=(2k+1)^2=4k^2+4k+1.$$ The former has the unique solution $k=4$, the latter has no solutions. This yields the solution $(a,b,p)=(11,7,2)$, corresponding to $$(m,n,p)=(3,1,2),$$ which we already found.