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user9072
user9072

Not sure this is good enough for an answer, but for $d=1$:

The polynomial $ax+ b = 0 $ has an integral root if an only if $a\mid b$. Let us ignore the case $b=0$ for now, and restrict to $a,b > 0$.

The number of couples $(a,b)$ with $1 \le a,b \le C$ such that $a\mid b$ can be expressed as $\sum_{1 \le n\le C} \tau(n)$ where $\tau(n)$ is the number of divisors of $n$.

It is known that $\sum_{1 \le n\le C} \tau(n) = C \log C + (2 \gamma - 1 ) C + O (\sqrt{C}) $ where $\gamma$ is the Euler–Mascheroni constant.

Thus among all $C^2$ polynomials with $1\le a,b \le C$ there are $C \log C + (2 \gamma - 1 ) C + O (\sqrt{C}) $ with an integral root, so the fraction is asymptotically $\log C / C$$\frac{\log C}{C}$.

And among all $2C(2C+1)$ polynomials wit degree $1$ with $-C \le a,b \le C$ there are $4(C \log C + (2 \gamma - 1 ) C + O (\sqrt{C})) + 2C+1$ with an integral root; the first term for the four combinations of signs and non-zero coefficients and then the $2C+1$ with constant term $0$ (or perhaps one should only count $2C$ not to count the zero-polynomials.)

So the asymptotic fraction is $\log C /C $$\frac{\log C}{C} $. This is not close for $C=100$, but taking the lower order terms into account it is not that bad.

Doing the actual calculation for $C=100$ one gets $2128$ non-zero polynomials with an integral root, for a fraction of around $5.29$ percent, quite close to the simulation.

Not sure this is good enough for an answer, but for $d=1$:

The polynomial $ax+ b = 0 $ has an integral root if an only if $a\mid b$. Let us ignore the case $b=0$ for now, and restrict to $a,b > 0$.

The number of couples $(a,b)$ with $1 \le a,b \le C$ such that $a\mid b$ can be expressed as $\sum_{1 \le n\le C} \tau(n)$ where $\tau(n)$ is the number of divisors of $n$.

It is known that $\sum_{1 \le n\le C} \tau(n) = C \log C + (2 \gamma - 1 ) C + O (\sqrt{C}) $ where $\gamma$ is the Euler–Mascheroni constant.

Thus among all $C^2$ polynomials with $1\le a,b \le C$ there are $C \log C + (2 \gamma - 1 ) C + O (\sqrt{C}) $ with an integral root, so the fraction is asymptotically $\log C / C$.

And among all $2C(2C+1)$ polynomials wit degree $1$ with $-C \le a,b \le C$ there are $4(C \log C + (2 \gamma - 1 ) C + O (\sqrt{C})) + 2C+1$ with an integral root; the first term for the four combinations of signs and non-zero coefficients and then the $2C+1$ with constant term $0$ (or perhaps one should only count $2C$ not to count the zero-polynomials.)

So the asymptotic fraction is $\log C /C $. This is not close for $C=100$, but taking the lower order terms into account it is not that bad.

Doing the actual calculation for $C=100$ one gets $2128$ non-zero polynomials with an integral root, for a fraction of around $5.29$ percent, quite close to the simulation.

Not sure this is good enough for an answer, but for $d=1$:

The polynomial $ax+ b = 0 $ has an integral root if an only if $a\mid b$. Let us ignore the case $b=0$ for now, and restrict to $a,b > 0$.

The number of couples $(a,b)$ with $1 \le a,b \le C$ such that $a\mid b$ can be expressed as $\sum_{1 \le n\le C} \tau(n)$ where $\tau(n)$ is the number of divisors of $n$.

It is known that $\sum_{1 \le n\le C} \tau(n) = C \log C + (2 \gamma - 1 ) C + O (\sqrt{C}) $ where $\gamma$ is the Euler–Mascheroni constant.

Thus among all $C^2$ polynomials with $1\le a,b \le C$ there are $C \log C + (2 \gamma - 1 ) C + O (\sqrt{C}) $ with an integral root, so the fraction is asymptotically $\frac{\log C}{C}$.

And among all $2C(2C+1)$ polynomials wit degree $1$ with $-C \le a,b \le C$ there are $4(C \log C + (2 \gamma - 1 ) C + O (\sqrt{C})) + 2C+1$ with an integral root; the first term for the four combinations of signs and non-zero coefficients and then the $2C+1$ with constant term $0$ (or perhaps one should only count $2C$ not to count the zero-polynomials.)

So the asymptotic fraction is $\frac{\log C}{C} $. This is not close for $C=100$, but taking the lower order terms into account it is not that bad.

Doing the actual calculation for $C=100$ one gets $2128$ non-zero polynomials with an integral root, for a fraction of around $5.29$ percent, quite close to the simulation.

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user9072
user9072

Not sure this is good enough for an answer, but for $d=1$:

The polynomial $ax+ b = 0 $ has an integral root if an only if $a\mid b$. Let us ignore the case $b=0$ for now, and restrict to $a,b > 0$.

The number of couples $(a,b)$ with $1 \le a,b \le C$ such that $a\mid b$ can be expressed as $\sum_{1 \le n\le C} \tau(n)$ where $\tau(n)$ is the number of divisors of $n$.

It is known that $\sum_{1 \le n\le C} \tau(n) = C \log C + (2 \gamma - 1 ) C + O (\sqrt{C}) $ where $\gamma$ is the Euler–Mascheroni constant.

Thus among all $C^2$ polynomials with $1\le a,b \le C$ there are $C \log C + (2 \gamma - 1 ) C + O (\sqrt{C}) $ with an integral root, so the fraction is asymptotically $\log C / C$.

And among all $2C(2C+1)$ polynomials wit degree $1$ with $-C \le a,b \le C$ there are $4(C \log C + (2 \gamma - 1 ) C + O (\sqrt{C})) + 2C+1$ with an integral root; the first term for the four combinations of signs and non-zero coefficients and then the $2C+1$ with constant term $0$ (or perhaps one should only count $2C$ not to count the zero-polynomials.)

So the asymptotic fraction is $\log C /C $. This is not close for $C=100$, but taking the lower order terms into account it is not that bad.

Doing the actual calculation for $C=100$ one gets $2128$ non-zero polynomials with an integral root, for a fraction of around $5.29$ percent, quite close to the simulation.