How many solutions does the equation $m^2-33n+1=0$, where $m,n\in\mathbb Z$, have?
The answer is no solutions exist. But why?
How many solutions does the equation $m^2-33n+1=0$, where $m,n\in\mathbb Z$, have?
The answer is no solutions exist. But why?
On
Look at the equation $\pmod 3$. The remainders of $m^2$ are $0$ or $1$, so any solution would give $0 + 0 + 1 \equiv 0 \pmod 3$ or $1 + 0 + 1 \equiv 0 \pmod 3.$
On
$n = \dfrac{m^2+1}{33}$. Observe that you can write $m = 33k+r$, and the problem boils down to finding those $r$ such that $0\leq r \leq 32$ with $33| r^2+1$.
On
Given, $m^2=33n-1$.
The digit sum of any multiple of $3$ is $3,6,9$.
(Here, digit sum is defined as the digit obtained by adding digits of given number until we get a single digit. For instance, digit sum of $3456$ is $9$.)
Also digit sum of a perfect square is $1$ or $4$ or $7$ or $9$.
So the possible digit sum of $m^2= 3-1=2$ or $6-1 =5$ or $9-1=8$. But we know that $m^2$ is perfect square, so the digit sum should be not be equal to $2$ or $5$ or $8$. So no solution exist.
Since every square is congruent to $0,1,3,4,9$ modulo $11$.