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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?

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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 $1 + 0 + 1 \equiv 0 \pmod 3.$

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Since every square is congruent to $0,1,3,4,9$ modulo $11$.

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  • $\begingroup$ Please clearly enplane it..... $\endgroup$
    – Empty
    Jan 14, 2015 at 12:01
  • $\begingroup$ Just consider all remainders when dividing by $11$ and square them. Specifically, $0^2\equiv 0$, $1^2\equiv 1$, $2^2\equiv 4$, $3^2\equiv 9$, $4^2\equiv 5$, $5^2\equiv 3$ $\endgroup$ Jan 14, 2015 at 12:08
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$n = \dfrac{m^2+1}{33}$. 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 $33| r^2+1$.

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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.

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