Food for thought

## Solution

Find the smallest positive integers $a$, $b$ and $c$ such that $a^2 = 2b^3 = 3c^5.$

Remember that we write $x \mid y$ to mean that $x$ divides $y$.

We say that two whole numbers are coprime if their highest common factor is $1$.

We require $a^2=2b^3$, so $2 \mid a^2$. Since $2$ is prime, we must have $2 \mid a$. Similarly, $a^2=3c^5$, so $3 \mid a^2$ and because $3$ is prime we must have $3 \mid a$.

Therefore, let $a=2^{\alpha_1}3^{\alpha_2}a_1$ where ${\alpha_1}$ and ${\alpha_2}$ are positive integers and $a_1$ is a positive integer coprime to $2$ and $3$.

When we substitute this representation of $a$ into the given equation, we obtain $2^{2\alpha_1}3^{2\alpha_2}a_1^2=2b^3=3c^5.$

Now we see that $2^{2\alpha_1} \mid 2b^3$, and because $\alpha_1$ is a positive integer there are at least two factors of $2$ in $2^{2\alpha_1}$. Therefore $b^3$ must be divisible by $2$, and because $2$ is prime, $2 \mid b$. Also, $3^{2\alpha_2} \mid 2b^3$, and because $3$ is prime we have $3 \mid b$.

Similarly $2 \mid c$ and $3 \mid c$, so let $b=2^{\beta_1}3^{\beta_2}b_1$ and $c=2^{\gamma_1}3^{\gamma_2}c_1$ where $\beta_1$, $\beta_2$, $\gamma_1$ and $\gamma_2$ are positive integers, and $b_1$ and $c_1$ are positive integers coprime to $2$ and $3$.

When we substitute these representations of $b$ and $c$—along with the representation of $a$ we used earlier—into the given equation, we obtain $2^{2\alpha_1}3^{2\alpha_2}a_1^2=2^{1+3\beta_1}3^{3\beta_2}b_1^3=2^{5\gamma_1}3^{1+5\gamma_2}c_1^5.$

We are using the uniqueness of prime factorisation here.

Since $3$, $a_1$, $b_1$ and $c_1$ are coprime to $2$ we must equate the powers of $2$, and similarly we must equate the powers of $3$.

So we must have $2\alpha_1=1+3\beta_1=5\gamma_1$ and $2\alpha_2=3\beta_2=1+5\alpha_2.$

Since $2\alpha_1=5\gamma_1$, we see that $5 \mid 2\alpha_1$ and because $2$ and $5$ are coprime we have $5 \mid \alpha_1$. So let $\alpha_1=5k$ where $k$ is a positive integer. Then $\gamma_1=2k$ and $1+3\beta_1=10k$. So $\beta_1=\frac{10k-1}{3}=3k+\frac{k-1}{3}.$

But $\beta_1$ must be a positive integer, so we must have $\frac{k-1}{3}=n$ where $n$ is a non-negative integer (notice that we can have $n=0$).

Then $k=3n+1$, and $\alpha_1=5(3n+1),$ $\beta_1=10n+3$ and $\gamma_1=2(3n+1).$

Using a similar process for $2\alpha_2=3\beta_2=1+5\gamma_2$ shows that $\alpha_2=3(5m+1),$ $\beta_2=2(5m+1)$ and $\gamma_2=6m+1$ for some non-negative integer $m$.

Therefore all solutions are of the form $a=2^{5(3n+1)}3^{3(5m+1)}a_1,$ $b=2^{10n+3}3^{2(5m+1)}b_1$ and $c=2^{2(3n+1)}3^{6m+1}c_1$ where $n$ and $m$ are non-negative integers, and $a_1$, $c_1$ and $b_1$ are positive integers coprime to $2$ and $3$.

To find the smallest possible solution we need to choose $a_1=b_1=c_1=1$ and $n=m=0$.

Then $a=2^53^3,$ $b=2^33^2$ and $c=2^23.$

What can you say about other solutions to this equation?

What can we say about $a_1$, $b_1$ and $c_1$ more generally?

Putting our representations of $a$, $b$ and $c$ into the given equation and cancelling all factors of $2$ and $3$ shows that $a_1^2=b_1^3=c_1^5.$

If $a$ has a prime factor $p$ then we can see immediately that $p$ divides $b$ and $c$. Similarly all prime factors of $b$ and $c$ are prime factors of $a$, $b$ and $c$. So the prime factorisations of $a$, $b$ and $c$ use exactly the same primes.

Let $\alpha$, $\beta$ and $\gamma$ be the greatest positive integers such that $p^\alpha \mid a$, $p^\beta \mid b$ and $p^\gamma \mid c$. For example, $p^\alpha \mid a$ but $p^{\alpha+1}$ does not divide $a$.

Then equating powers of $p$ in the given equation shows that $2\alpha=3\beta=5\gamma.$

This has solution $\alpha=15r,$ $\beta=10r$ and $\gamma=6r$ where $r$ is a positive integer.

Therefore we can let $a_1=d^{15},$ $b_1=d^{10}$ and $c_1=d^6$ where $d$ is any positive integer coprime to $2$ and $3$.

This gives the general solution $a=2^{5(3n+1)}3^{3(5m+1)}d^{15},$ $b=2^{10n+3}3^{2(5m+1)}d^{10}$ and $c=2^{2(3n+1)}3^{6m+1}d^6$ where $n$ and $m$ are any non-negative integers and $d$ is any positive integer coprime to $2$ and $3$.

Notice this can be rearranged to give $a=2^53^3(2^n3^md)^{15},$ $b=2^33^2(2^n3^md)^{10}$ and $c=2^23(2^n3^md)^6$ so it is easy to see where our smallest solution came from.