# Show that there are no two positive integers verifying this equality

I'm trying to show that there is no pair of strictly positive integers $\alpha$ and $\beta$ such that $$180n^2+218n+66 = \alpha (30n+18) + \beta(36n+22)$$ but so far I haven't quite managed to do it. Any ideas?

Thanks

• A basic idea would be to study the polynomial $180 n^{2} + (218 -30\alpha -36\beta)n +66 -18\alpha -22\beta$ , what are its roots ? is it possible that it admits roots if $\alpha$ and $\beta$ are positive integers ? – krirkrirk Mar 7 '15 at 18:21
• You probably need to use the fact that $(30n+18)(36n+22) = 6(180n^2+218n+66)$. – Yuval Filmus Mar 7 '15 at 18:24

we have $5(36n+22)-6(30n+18)=2$.so we have $(36n+22,30n+18)=2$ and $$180n^2+218n+66=5(90n^2+109n+33)(36n+22)-6(90n^2+109n+33)(30n+18)$$ so $$\beta=5(90n^2+109n+33)+(15n+9)t,\alpha=-6(90n^2+109n+33)-(18n+11)t$$ for $t=-36(36+22)$ we have $\alpha=0$ and yo can see $\beta<0$ for t-1