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For $T={\sum }_{a=1}^{p-1}a(\frac{a}{p})$, one has $T=0$ when $p\equiv 1(\mathrm{mod}4)$, and $T\equiv 0(\mathrm{mod}p)$ for every $p>3$.

For an odd prime $p$,For $p\nmid a$, write $a=g^m$ using a primitive root $g$; then the left side is $(-1{)}^m$, which is one exactly when $m$ is even and $a$ is a square.

For an odd prime $p$ and $p\nmid a$, Euler's criterion identifies the Legendre symbol by modular exponentiation:

Thus a quadratic residue gives one and a quadratic nonresidue gives minus one, represented by $p-1$ if a nonnegative remainder is used. The formula also holds when $p\mid a$, because $(p-1)/2>0$ and both sides are zero modulo $p$.^[ref]

For example, modulo seven the exponent is three: $2^3\equiv 1$ confirms that two is a quadratic residue, while $3^3\equiv 6\equiv -1$ confirms that three is a quadratic nonresidue. Modulo eleven the exponent is five, and $2^5=32\equiv 10\equiv -1$ confirms that two is a quadratic nonresidue. The test answers whether a square root exists; it does not itself return the roots $3,4$ of $x^2\equiv 2(\mathrm{mod}7)$.

A short proof needs only Fermat's little theorem and the root bound for a polynomial. Put $m=(p-1)/2$. If $a=x^2\ne 0$ in the prime field, then $a^m=x^{p-1}=1$. There are already $m$ such quadratic residues. The nonzero polynomial $T^m-1$ has polynomial degree $m$, so it has no other polynomial roots. Finally, for every nonzero $a$, Fermat's little theorem gives $(a^m{)}^2=a^{p-1}=1$. Factoring $(a^m-1)(a^m+1)=0$ in the field shows $a^m\in \{1,-1\}$; every quadratic nonresidue must take the remaining value $-1$. This proves the criterion without first finding a primitive root.

For a larger original example, decide whether seven is a quadratic residue modulo the prime number $101$. Exponentiation by squaring computes the required power without listing fifty squares:Therefore $(7/101)=-1$ and $x^2\equiv 7(\mathrm{mod}101)$ has no solution. Each intermediate product is reduced modulo $101$, as in modular exponentiation. The advantage is that exponentiation by squaring uses a number of multiplications proportional to the binary digit length of the exponent rather than testing every potential square root.

The odd prime hypothesis is essential. For the composite modulus nine, $2^{(9-1)/2}=2^4\equiv 7(\mathrm{mod}9)$, which is neither one nor minus one. Although four is a nonzero square modulo nine, $4^4\equiv 4(\mathrm{mod}9)$, so even the square-to-one implication fails. The Jacobi symbol and Euler pseudoprime discuss different questions for odd composite moduli; they do not make this a universal square test.

If a smooth self-map of a compact manifold has only isolated nondegenerate fixed points, thenEvery summand has absolute value one, so the number of fixed points is at least $|L(f)|$.

For a bounded open $U\subset {\mathbb{R}}^n$, $m^2>0$, and $f\in L^2(U)$, there is a unique $u\in H_0^1(U)$ such thatfor every $v\in H_0^1(U)$.

Let a positive-measure part ${\Gamma }_D$ of the boundary of a bounded connected domain carry a homogeneous Dirichlet boundary condition, and let the complementary part carry a homogeneous Neumann boundary condition. Onthe weak Poisson problem isThe Poincare inequality with a partial Dirichlet boundary makes the left side coercive, so the Lax-Milgram theorem gives a unique weak solution.

For $\ddot{x}=-U^{\prime }(x)-\mu \dot{x}$ with $\mu >0$,If $U$ is radially unbounded, every trajectory is bounded, and LaSalle's principle reduces its omega-limit set to invariant points with zero velocity.

The Heaviside function is zero before its switching time and one after it, allowing delayed signals to be written algebraically.

The delayed unit step has

After $t=zs$, the phase in the integral for ${\Gamma }^{\prime }(z)$ becomes $\log s-s$, with maximum $-1$ and second derivative $-1$ at $s=1$. Hence

A Laplacian eigenfunction satisfies $-\Delta u=\lambda u$ together with specified boundary conditions.

The spherical-coordinate Laplacian splits into a radial operator and the sphere’s angular Laplacian divided by $r^2$.

For radial $u(r)$ in $n$ dimensions, $\Delta u=u\prime \prime +(n-1)u\prime /r$.

Laplace's equation isIts solutions are harmonic functions.

If a twice differentiable real function has a local maximum at an interior point of a Riemannian manifold, its Hessian matrix is negative semidefinite there and hence its Laplacian is nonpositive.

A tricritical point is where a line of continuous transitions meets a line of first-order transitions; in a ${\mathbb{Z}}_2$ Landau expansion both quadratic and quartic coefficients vanish and a positive sextic term stabilizes the free energy.

For an ordinary quartic Landau transition, the heat-capacity and order-parameter exponents are $\alpha =0$ and $\beta =1/2$. At a tricritical sextic point they are $\alpha =1/2$ and $\beta =1/4$.

Landau-Ginzburg theory supplements the local Landau potential with gradient terms and treats the order parameter as a spatial field.

Stokes drift is the difference between the mean velocity of fluid particles and the Eulerian mean velocity at fixed positions. It appears at second order because particles sample the oscillatory velocity at displaced positions.

For a field $\Phi$,