The original problem is: "Find all possible pairs of positive integers $(a, b)$

$$k = \dfrac{a^3+300^2}{a^2b^2+300}\tag1$$

such that $k$ is an integer."

I've tried so many different ways. Now this question comes up from one of them. Let,

$$\left(\dfrac{a}{10} \right)^2=\dfrac{3(300-k)}{kb^2-a}=\dfrac{p^2}{q^2}\tag2$$

If you calculate $k$ in terms of $a, b, p, q,$ then the following question comes up:

"Suppose $a$, $b$, $p$ and $q$ are natural numbers such that $a<300$ and $\gcd(p, q)=1$. Is it possible for

$$k=\frac{900q^2+ap^2}{3q^2+b^2 p^2}\tag3$$

to become an integer?"

I think this is easier than original problem, but I don't know how to proceed from here.

  • $\begingroup$ @TitoPiezasIII yeah, regarding given conditions it has three solutions. $\endgroup$ – Ghartal Sep 7 '15 at 7:24
  • $\begingroup$ @PatrickStevens you are right, we can use $a<300$ only. Let me edit the problem. $\endgroup$ – Ghartal Sep 7 '15 at 7:25
  • $\begingroup$ You made a typo somewhere as there are many more solutions than that. What are the three you found? $\endgroup$ – Tito Piezas III Sep 7 '15 at 7:25
  • $\begingroup$ Solutions are $(a, b)=(10, 2), (20, 1), (30, 2)$. $\endgroup$ – Ghartal Sep 7 '15 at 11:20

While your $(3)$ can be derived from $(2)$, if it is an integer does not guarantee that $(1)$ will be an integer as well. Note that it has the four variables $a,b,p,q$ that can "integerize" your expression, while you're stuck with only $a,b$ for $(1)$.

(For example, $a,b,p,q =3,2,9,5$ makes $(3)$ an integer, but $a,b = 3,2$ does not for $(1)$.)

Also, your original problem should explicitly specify the constraint that $a \neq 300b^2$. If not, then $a = 300b^2$ yields an infinite family which always has $k=300$.

A computer search with $a<10000$ and $b<100$ yields the infinite family and,

$$a,b,k = 10,2,130$$

$$a,b,k = 20,1,140$$

$$a,b,k = 30,2,30$$

which seem to be the only other solutions.

  • $\begingroup$ Thanks Tito. It can be proven that $a<300$ and doing a computer search gives the answers you have provided. So these are the only solutions other than the case $a=300b^2$, but I can't prove it by hand. $\endgroup$ – Ghartal Sep 7 '15 at 18:33

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