Tuymaada 2015, Day 2, Problem 6, Senior League states:
Consider integers $a,b,c,d$ such that $0 \leq b \leq c \leq d \leq a$ and $a>14$. Prove that there exists a positive integer $n$ that can not be represented as $$n=x(ax+b)+y(ay+c)+z(az+d)$$for any integers $x,y,z$. (K. Kohas)
The expression can be rewritten as
$$a(x^2 + y^2 + z^2) + bx + cy + dz = n.$$
Obviously $gcd(a,b,c,d)$ divides $n$, so if $gcd(a,b,c,d) > 1$ we can just choose $n$ a nonmultiple of $gcd(a,b,c,d)$ and we're done.
Thus the problem now boils down to showing that if $gcd(a,b,c,d) = 1$ and $(x,y,z,t)$ satisfy $$at + bx + cy + dz = 1,$$ then $t \neq x^2 + y^2 + z^2$.
How could I prove this? I have no idea where the condition $a > 14$ fits into the picture. I tried writing $b = \alpha, c = \alpha + \beta, d = \alpha + \beta + \gamma, a = \alpha + \beta + \gamma + \delta$ such that $$\alpha(x^2 + y^2 + z^2 + x + y + z) + \beta(x^2 + y^2 + z^2 + y + z) + \gamma(x^2 + y^2 + z^2 + z) + \delta(x^2 + y^2 + z^2) = n.$$
Now the condition becomes just $\alpha + \beta + \gamma + \delta > 14$, where $\alpha, \beta, \gamma, \delta$ are integers $\geq 0$. It is easy to show that if $gcd(a,b,c,d) = 1$ then $gcd(\alpha, \beta, \gamma, \delta) = 1$. I could not get anything out of this approach, but it might be helpful.
Note that $ax^2+bx\ge 0$ and similarly for the other terms. Note also that because of $a\ge b$ this expression is at least $2a\ge 30$ if $|x|\ge2$ and similarly for the other terms. So for any $n<30$, we have to choose $x,y,z$ among $0,-1,1$, but this gives at most $27$ different values so there is an $n$ that will not be reached.