Solution of A Quintic Equation with Exactly One Real Solution

Step 0: the ambient interval. $\mathbb{R}$ is an interval, and every $x\in\mathbb{R}$ is an interior point of it, as recorded there.

Step 1: $p$ is continuous on $\mathbb{R}$. Since $p$ is a polynomial function on $\mathbb{R}$, its restriction to any subset of $\mathbb{R}$ is continuous there by The Real Line: Standing Notation and Background for Calculus on Intervals (v); in particular $p$ is continuous on $\mathbb{R}$.

Step 2: $p$ is differentiable with $p'(x)=5x^4+1$. Let $x\in\mathbb{R}$. By clause 1 of Derivative of a Polynomial Function on the Real Line, the map $z\mapsto z$ is differentiable at $x$ with derivative $1$, and the map $z\mapsto z^5$ is differentiable at $x$ with derivative $5x^4$. By clauses 1 and 2 of Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (constants and sums), $p$ is differentiable at $x$ with

$$p'(x)=5x^4+1 .$$

Step 3: $p$ is strictly increasing on $\mathbb{R}$. Let $x\in\mathbb{R}$. By clause 1 of Properties of Natural Number Powers in a Field applied repeatedly, $x^{2}=x\cdot x$, $x^{3}=x^{2}\,x$ and $x^{4}=x^{3}\,x$; by associativity and commutativity of multiplication these give

$$x^{4}=(x\cdot x)(x\cdot x),$$

which is a square, so $0\le x^{4}$ by Nonnegativity of Squares in an Ordered Field. Now $0<5$, since $0<1$ by clause 6 of Elementary Order Arithmetic in an Ordered Field and repeated use of clause 3 of that lemma adds $0<1$ to itself. If $x^{4}=0$ then $5x^{4}=0$; otherwise $0<x^{4}$ and clause 5 gives $0<5x^{4}$. In either case $0\le 5x^{4}$. Applying clause 3 to $0<1$ and $0\le 5x^{4}$ gives

$$0<5x^4+1=p'(x) .$$

Thus $p$ is continuous on the interval $\mathbb{R}$ and differentiable at every interior point of it, with everywhere positive derivative. By clause 2 of The Sign and Size of the Derivative: Monotonicity, Constancy and Lipschitz Bounds, $p$ is strictly increasing on $\mathbb{R}$.

Step 4: at most one zero. Let $c,c'\in\mathbb{R}$ with $c\ne c'$. Since the order of $\mathbb{R}$ is total, either $c<c'$ or $c'<c$; in the first case $p(c)<p(c')$ and in the second $p(c')<p(c)$, so in both cases $p(c)\ne p(c')$. Hence at most one real number $c$ satisfies $p(c)=0$.

Step 5: at least one zero, lying strictly between $0$ and $1$. We compute

$$p(0)=0+0-1=-1,\qquad p(1)=1+1-1=1 ,$$

so $p(0)\le 0\le p(1)$. The restriction of $p$ to $[0,1]$ is continuous on $[0,1]$ by clause 1 of Restriction Stability of Continuity and of the Derivative, and $0<1$. By Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals applied with the value $0$ there is $c\in[0,1]$ with $p(c)=0$. Since $p(0)=-1\ne 0$ and $p(1)=1\ne 0$, we have $c\ne 0$ and $c\ne 1$, so

$$0<c<1 .$$

Steps 4 and 5 together show that there is exactly one real number $c$ with $p(c)=0$, and that $0<c<1$.