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$m, n,p \in \mathbb{N}^*$ such that $m+n+p = 2017$.

Find the values of $x,y,z,t$ Such that $x+my+nz+pt =2018$ With $x,y,z,t \in \mathbb N^*$

My Attempt:

Im not sure if this solution is correct because it looks easy to me.

If we add a $1$ to the original equation: $$1+1(m)+(1)n+(1)p=2018$$

And fron here, It’s easy to see that the only solution to the equation is $(x,y,z,t)=(1,1,1,1)$, Given the fact that $(x,y,z,t) \in \mathbb N^*$ so they cannot equal to $0$.

PNT
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1 Answers1

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Yes. You can formalize this by considering $x=1$ and $x\geq 2$ by observing that $$x+m(y-1)+n(z-1)+p(t-1) + m + n+ p =2018 \\ \implies x+m(y-1)+n(z-1)+p(t-1) =1$$

$x=1$: $$ m(y-1)+n(z-1)+p(t-1)=0 \\ \implies (y-1)=(z-1)=(t-1) = 0 \\ (y,z,t) = (1,1,1) $$

$x\geq 2$: $$ m(y-1)+n(z-1)+p(t-1)\lt 0 $$ so at least one of $(y{-}1),(z{-}1),(t{-}1)$ is less than zero and the corresponding in $y,z,t$ is not $\in \mathbb N^*$

For style you could perhaps define $y',z',t'\in \mathbb N_0$ to be the decremented versions of $y,z,t$

@zwim in comments points out that we can extend the initial reformulation: \begin{align} x+m(y-1)+n(z-1)+p(t-1) + m + n+ p &=2018 \\ \implies x+m(y-1)+n(z-1)+p(t-1) &=1 \\ \implies (x-1)+m(y-1)+n(z-1)+p(t-1) &=0 \\ \implies (x-1)=(y-1)=(z-1)=(t-1) &=0 \end{align}

giving us $(x,y,z,t) = (1,1,1,1)$ as stated.

Joffan
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    Why separating in two cases ? I would state directly $(x-1)+m(y-1)+n(z-1)+p(t-1)=0\implies (x-1)=(y-1)=(z-1)=(t-1)=0$. – zwim Feb 12 '21 at 19:03
  • @zwim Yes, that works nicely. I just wanted to emphasize the lack of solutions with larger $x$. – Joffan Feb 12 '21 at 19:09
  • @joffan did you knew about the answer before or this is the first time you see this problem? – PNT Feb 12 '21 at 22:07
  • @Yassir - First time I saw it, but since you had pretty much solved it in the question it wasn't that hard to tidy up the proof. – Joffan Feb 12 '21 at 22:49