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.
