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The question read "show that $19m^2+95mn+2000n^2=1995$ has no integer solution for $n$ and $m$." I have attempted a solution and would like to check if it is correct.

$95mn +2000n^2 = 1995-19m^2$ now factorize both sides $5n(19m+400n) = 19(105-m^2)$ From here I tried all parity cases for $m$ and $n$ and constantly end up with $odd = even$ or $even = odd$.

Is this correct? Thanks.

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It's rather hard to tell whether your argument is correct or not without being given the actual argument. Telling us you used a parity argument doesn't say much. In particular, I don't even see how to proceed with a parity argument. – EuYu Feb 17 '13 at 9:00
Start with, m and n are odd, then m and n are even, then m is odd m is even, m is even n is odd. – fosho Feb 17 '13 at 9:02
What exactly is wrong with $n$ even and $m$ odd? – EuYu Feb 17 '13 at 9:03
Parity consideration is not enough as $n$ even and $m$ odd doesnt give a contradiction. – Ivan Loh Feb 17 '13 at 9:07
up vote 5 down vote accepted

Clearly, $19\mid n,n=19r$(say)

So, $$19m^2+95mn+2000n^2=1995\implies 19m^2+95m(19r)+2000(19r)^2=1995$$

So, $$m^2+95mr+2000(19)r^2=105\implies m^2\equiv10\pmod {19} $$

But $10$ is not a Quadratic residue of $19$

as $(\pm1)^2\equiv1\pmod{19},(\pm2)^2\equiv4,(\pm3)^2\equiv9,$ $(\pm4)^2\equiv16,(\pm5)^2=25\equiv6,(\pm6)^2=36\equiv17,$ $(\pm7)^2=49\equiv11,(\pm8)^2=64\equiv7,(\pm9)^2=81\equiv5$

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Taking $\pmod 5$ gives $5 \mid m$. Now taking $\pmod {25}$ gives $25\mid 1995$, a contradiction.

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If $19m^2+95mn+2000n^2 = 1995$ then reducing mod $5$ tells us that $m^2 \equiv 0 \bmod 5$, i.e. that $5|m$.

But then reducing both sides mod $25$ gives $0 \equiv 20 \bmod 25$ which is a contradiction.

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Show that $5\mid m$ therefore $25\mid 19m^2+95mn+2000n^2$. Since $25\nmid 1995$ the result follows.

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$\rm\begin{eqnarray}{\bf Hint}\ \ \ (p,a)\color{#C00}=1,\ \, and\, \ \ \color{#0A0}{squarefree}\ \ p\mid ax^2+bpx + cp^2\\ \rm \Rightarrow p\mid ax^2\ \color{#C00}\Rightarrow\ \rm p\mid x^2\ \color{#0A0}{\Rightarrow}\ \rm p\mid x\ \Rightarrow\ p^2\mid ax^2+bpx + cp^2\end{eqnarray}$

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