Leicich wrote: A pair of positive integers $(a,b)$ is called charrua if there is a positive integer $c$ such that $a+b+c$ and $a\times b\times c$ are both square numbers; if there is no such number $c$, then the pair is called non-charrua. a) Prove that there are infinite non-charrua pairs. b) Prove that there are infinite positive integers $n$ such that $(2,n)$ is charrua. a) $(1,(2k+1)^2)$ is non-charrua because $1+(2k+1)^2+c^2=2,3\mod 4$ can not be square. b) $(2,16k^2-2)$ is charrua because we can let $(x,y)$ be solution to $x^2-(8k^2-1)y^2=1$, then pick $c=16k^2(x^2-1)$.

For part a) all pairs $(a,b)$ with $a \equiv b \equiv 1 \pmod 4$ are non-charrua. For part b) we can also take $n = 2(2k+1)^2$. The corresponding value of $c$ would be $c = (2k^2+2k)^2$. It is easy to verify that in this case $2+n+c = (2k^2 + 2k+2)^2$. My solution for part b) comes from noting that by letting $n$ be of the form $2p^2$ then $c$ is forced to be a perfect square, which makes solving the equation $2+n + c = x^2$ easier.

a) We claim that $(m,m)$ is non-charrua for all odd $m$. Clearly, for it to be charrua, we need $c$ to be a perfect square. But then $2m+c$ can only be $2$ or $3$ congruent $\pmod 4$, which means it cannot be a square. b) We claim that $(2,2m^2)$ is charrua for all odd $m$. Indeed, if we let $c=\frac{1}{4}(m^4-2m^2+1)$, we have that both $$2\cdot 2m^2\cdot \frac{1}{4}(m^4-2m^2+1) = (m^3-1)^2$$and $$2+2m^2+\frac{1}{4}(m^4-2m^2+1)=\left(\frac{m+3}{2}\right)^2$$are perfect squares.