Does $ 18x + 7y ≡ 0 \pmod {m^2} \implies \max\{x,y\}>\frac{m^2}{25}$?

Question

Let, $M=\max\{x,y\}$. I found following -

we have - $ 18x + 7y ≡ 0 \pmod {m^2}.$ Then $ M ≥ m^2/25 $.

The only possible way seems to be is that $ 18x + 7y ≡ 0 \pmod {m^2} \implies $ either $x$ or $y$ is divided by $m^2 \implies M=m^2k \implies M>\frac{m^2}{25}$.

But I can't figure out how $ 18x + 7y ≡ 0 \pmod {m^2} \implies $ either $x$ or $y$ is divided by $m^2$, Can any one explain?

No; $18(1)+7(1)\equiv0\mod 5^2$ but $1$ is not divisible by $5^2$ — J. W. Tanner, Jul 22 '19 at 13:51
@J.W. Tanner, do you want this one, or should I try clearing things up ? — , Jul 22 '19 at 13:54
Please undelete your latest question (else those who may have been working on answers will have wasted their time) — Bill Dubuque, Aug 03 '19 at 13:37

score 4 · Answer 1 · answered Jul 22 '19 at 13:52

4

This has nothing to do with modular arithmetic. Rather, if $x,y$ are positive, and $18x+7y\geq m^2$ (which happens if $18x+7y\equiv_{m^2}0$), then $\max(x,y)\geq \frac{m^2}{25}$.

Proof: Assume, for contradiction, that they are both smaller than $\frac{m^2}{25}$. Then we get $$ 18x+7y<18\frac{m^2}{25}+7\frac{m^2}{25}=m^2 $$

answered Jul 22 '19 at 13:52

Arthur

199,419

1

Sorry about the OP changing acceptance. – Jul 22 '19 at 15:27
4

@RoddyMacPhee That's nothing to apologize for. If the OP thinks your answer works better for them, then changing the acceptance is their prerogative. If anything, they may have been too quick to give it out in the first place, as waiting a bit is more inviting for deeper, longer answers. – Arthur Jul 22 '19 at 15:29

Does $ 18x + 7y ≡ 0 \pmod {m^2} \implies \max\{x,y\}>\frac{m^2}{25}$?

1 Answers1