Why $n\geq 50$? Among the natural numbers $n<50$, the largest not so expressible is $23$.

mavropnevma wrote: Why $n\geq 50$? Among the natural numbers $n<50$, the largest not so expressible is $23$. It was that so we wouldn't need to consider small cases.

Let $n= t^2 + s$, where $ 0\leq s \leq 2t $, that's $ t \leq \sqrt{n}<t+1$. For $s=0$ we have the representation: $n=t(t-1)+t$. For $1\leq s \leq t$, we have $n=t^2 + s$. Suppose $t+1$ is not a prime. Then, for $t<s\leq 2t$ ,we have $n= t(t+1)+(s-t)$. Let us suppose, for now on, $t+1$ is a prime number. In that case, $t+2$ is not a prime. For $t<s\leq 2t-2$, we have $n=(t-1)(t+2)+ (s-t+2)$. When $s=2t$, we have $n=t^2+2t$ It remains the case $s=2t-1$. It that case, $n=(t-2)(t+4) + 7$, Since $7<\sqrt{n}$, the last representation is OK.

dgrozev wrote: Let $n= t^2 + s$, where $ 0\leq s \leq 2t $, that's $ t \leq \sqrt{n}<t+1$. For $s=0$ we have the representation: $n=t(t-1)+t$. For $1\leq s \leq t$, we have $n=t^2 + s$. Suppose $t+1$ is not a prime. Then, for $t<s\leq 2t$ ,we have $n= t(t+1)+(s-t)$. Let us suppose, for now on, $t+1$ is a prime number. In that case, $t+2$ is not a prime. For $t<s\leq 2t-2$, we have $n=(t-1)(t+2)+ (s-t+2)$. When $s=2t$, we have $n=t^2+2t$ It remains the case $s=2t-1$. It that case, $n=(t-2)(t+4) + 7$, Since $7<\sqrt{n}$, the last representation is OK. Case for $1\leq s \leq t$??????

dgrozev wrote: ... For $1\leq s \leq t$, we have $n=t^2 + s$... This is the needed representation in case $1\le s\le t.$

Is the example right in the question? As 3 does not divide 80 as well as 14.