The following three theorems are familiar and commonly used mathematical principles in algebraic operations, and will not be proven here.

Theorem 2. 1. If a, b and c are integers, there must be $a+b=c$.

Theorem 2.2: If a, b and c are integers and $a+b=c$ holds, then $\left(a+b\right)d=cd$ also holds.

Theorem 2.3. If a, b and c are integers, $ab=c$, Then c must contain factor a or b.

In this article, we will apply the following congruences theory, for example

If $a\equiv c\left(\mathrm{mod}p\right)$, $b\equiv d\left(\mathrm{mod}p\right)$, and $e=c+d$,

then

$a+b\equiv e\left(\mathrm{mod}p\right)$.

If $c+d=0$, then $a+b\equiv 0\left(\mathrm{mod}p\right)$.

Regarding the theory and formulas of congruence, we can refer to some books [4] .

Factorization method: It is applicable to polynomial equations, characterized in that all equations are in the following form

$k_1{x}^i+k_2{x}^{i-1}+\cdots +k_{i}x+f=0$ (1)

where $k_1,k_2,\cdots ,k_i$ and $f$ are integers.

So, the equation (1) can be changed to

$\left(k_1{x}^{i-1}+k_2{x}^{i-2}+\cdots +k_i\right)x=-f$. (2)

It is clear that the condition for equation (2) to hold is $x|f$ or $\left(k_1{x}^{i-1}+k_2{x}^{i-2}+\cdots +k_i\right)|f$ [5] .

Example 3.1: Attempt to find an integer solution to the quadratic equation (3) of one variable

$x^4+x^3+x^2+x-26520=0$. (3)

The solution can be obtained from equation (4)

$x^4+x^3+x^2+x=26520$, (4)

$\left(x^3+x^2+1\right)x=26520$. (5)

According to the Fundamental Theorem of Arithmetic and Theorem 3, assuming equation (5) has an integer solution, 26520 must contain a factor x. Given that 26,520 = 5 × 8 × 13 × 51, x may be one of them. Now, substitute each factor into equation (5) and test them one by one. When $x=-13$, substituting it into the equation yields

${\left(-13\right)}^4+{\left(-13\right)}^3+{\left(-13\right)}^2+\left(-13\right)=26520$,

$28561-2197+169-13=26520$

Obviously, when $x=-13$, both sides of equation (3) are equal, so $x=-13$ is the solution to equation (3).

This method is applicable to equations of the following form

$A^x+B^y+\cdots +U^v=W^z$. (6)

The condition for using the Index method in equation (6) is that all exponents on the left side of the equation are coprime with z. Additionally, the left-hand side of the equation can have coefficients.

We refer to the equation in the form of equation (7) as the Beal equation [6] . Below we will introduce its solution.

$A^x+B^y=C^z$. (7)

where $A,B,C,x,y,z$ are integers.

Regardless of whether a, b, x, and y are any integers, an integer c is can be found to make equation (8) hold (equation (8) is called the initial equation).

$a^x+b^y=c$ (8)

By Theorem 3.2.1, multiply both sides of the initial equation (8) by $c^{xy}$ to obtain

$\left(a^x+b^y\right)c^{xy}=c{c}^{xy}$,

$a^x{c}^{xy}+b^y{c}^{xy}=c{c}^{xy}$. (9)

Rewrite equation (9) as

${\left(a{c}^y\right)}^x+{\left(b{c}^x\right)}^y=c^{xy+1}$.

So, if $z=xy+1$ or $z|\left(xy+1\right)$, it is easy to see that the solution to equation (7) is $A=a{c}^y$, $B=a{d}^y$, $C=c$..(If $z|\left(xy+1\right)$, $xy+1=kz$, then $C=c^k$).

Example 3.2.1 Solving the Beal equation

$A^3+B^5=C^4$. (10)

According to the above method, for equation (10), we need to establish a new equation

$a^3+b^5=c$. (11)

Foe equation (11), let a = 3, b = 2, and substituting them into the left side of equation (11) yields a new equation (12)

$3^3+2^5=59$, (12)

Multiply ${59}^{3\times 5}$ on both sides of equation (12) to obtain

$\left(3^3+2^5\right){59}^{3\times 5}=59\times {59}^{3\times 5}$,

$3^3\times {59}^{3\times 5}+2^5\times {59}^{3\times 5}=59\times {59}^{3\times 5}$, (13)

Equation (13) can be rewritten as:

${\left(3\times {59}^5\right)}^3+{\left(2\times {59}^3\right)}^5={\left({59}^4\right)}^4$, (14)

Obviously, the solution to equation (10) is $A=3\times {59}^5=714924299$, $B=2\times {59}^3=410758$, and $C={59}^4=12117361$.

The third algorithm is applicable to multivariate multiple equations, characterized in that all equations as equation (15) (Simply put, it contains a term with an unknown exponent of 1)

$k_1{A}^i+k_2{B}^{i-1}+\cdots +k_2{U}^2+k_{i}V+n=0$, (15)

where $k_1,k_2,\cdots ,k_i$ and $n$ are integers.

Its L-algorithm is to transform equation (15) into

$k_1{A}^i+k_2{B}^{i-1}+\cdots +k_2{U}^2+n=-k_{i}V$ (16)

Let

$w=k_1{A}^i+k_2{B}^{i-1}+\cdots +k_2{U}^2+n$,

then by equation (16) we obtain

$V=-w/k_i$.

1) For the rational number solution of equation (Generol method);

Example 3.3.1. Try to find the rational number solution of the fifth degree equation (18).

$3a^5+7b^4+11c^3+17d^2+23e+71=0$. (18)

Let

$w=3a^5+7b^4+11c^3+17d^2+71$. (19)

Since only seeking that the solution to the equation is a rational number, the unknowns on the left side of the equation can be any integer (of course, the smaller the better), where w is an integer and V is a rational number.

For example, let a = 3, b = 4, c = 5, d = 2, substitute them into the left part of equation (19) to obtain

$w=3\times 3^5+7\times 4^4+11\times 5^3+17\times 2^2+71=4035$,

$e=-\frac{w}{23}=-\frac{4035}{23}$.

In this way, the rational number solution of equation (13) is a = 3, b = 4, c = 5,

d = 2. $e=-\frac{4035}{23}$.

2) The solution to the equation are integers, which can be divided into two situations:

a) When $k_1|n$:

$k_1{A}^i+k_2{B}^{i-1}+\cdots +k_2{U}^2+k_{i}V+n=0$ (20)

where $k_1,k_2,\cdots ,k_i$ and $n$ are integers, and $k_1|n$.

Since the solution to the equation is an integer and $k_1|n$,the unknowns $A,B,\cdots ,U$ on the left side of the equation can be chosen as any multiple of $k_i$ (of course, the smaller the better), then

$w=k_1{A}^i+k_2{B}^{i-1}+\cdots +k_2{U}^2+n$, $w\equiv 0\left(\text{mod}{k}_i\right)$;

and $V=-w/k_i$ ( $V$ is an integer).

In this way, the equation (20) can obtain a set of integers of a solution.

Example 3.3.2. Try to find the integer solution of the fifth degree equation (21).

$3a^3+7b^2+9c+81=0$. (21)

Since 9|81, set a = 9, b = 18, substitute them into the equation (21) to obtain

$3\times 9^3+7\times {18}^2+9c+81=0$

$4455+9c=0$,

$c=-495$.

So, the solution of the equation (20) are a = 9, b = 18, c = −495.

Or, set a = 3, b = 3, substitute them into the equation (21) to obtain

$3\times 3^3+7\times 3^2+9c+81=0$

$225+9c=0$,

$c=-25$.

So, the other solution of the equation (20) are a = 3, b = 3, c = −25.

b) When here is not $k_1|n$ (need congruence algorithm) [6] ;

Example 3.3.3. Try to solving equation (22)

$4a^5+7b^4+11c^3+17d^2+19e+791=0$. (22)

Since

$w=4a^5+7b^4+11c^3+17d^2+791$, (23)

in order to make

$w\equiv 0\left(\mathrm{mod}19\right)$,

we can use the theory of congruence to obtain the values of a, b, c and d, for example:

Let $a=19$, $4a^5=4\times {19}^5=9904396$, $9904396\equiv 0\left(\mathrm{mod}19\right)$;

Let $b=1$, $7b^4=7\times 1^4=7$, $7\equiv 7\left(\mathrm{mod}19\right)$,

Let $c=3$, $11c^3=11\times 3^3=297$, $297\equiv 12\left(\mathrm{mod}19\right)$,

Let $d=5$, $17d^2=17\times 5^2=425$, $425\equiv 7\left(\mathrm{mod}19\right)$,

and

$791\equiv 12\left(\mathrm{mod}19\right)$;

so

$0+7+12+7+12\equiv 0\left(\mathrm{mod}19\right)$.

Now, we have found that are a = 19, b = 1, c = 3, and d = 5, substitute them into the equation (23) to obtain

$\begin{array}{c}w=4a^5+7\times b^4+11\times c^3+17\times d^2+791\\ =4\times {19}^5+7\times 1^4+11\times 3^3+17\times 5^2+791\\ =9905916\end{array}$

Since by the equation (22) we obtain

$w+19e=0,$

then

$19e=-w=-9905916,$,

$e=-9905916/19=-521364$.

In this way, the integer solution of equation (23) is a = 19, b = 1, c = 3, d = 5, e = −521364.

Definition 2: The above new algorithms for equations are referred to as L-algorithms.

1) The L-algorithm only uses multiplication and addition/subtraction (summing up some power terms), and finally uses division to solve. Because multiplication is easier than finding square roots and can be manually operated, this algorithm has significant superiority and operability. In theory, it is not limited by pluralism and large power exponents.

2) By utilizing the principle of L-algorithm, we can also solve more similar types of algebraic equations. for example equation (24)

Example 4.1. Try to solving equation (24)

$A^7+7B^2+17C+89=0$ (24)

Because equation (24) has an unknown whose exponent is one, we can use the Congruent complementarity method, as let

$w=A^7+7B^2+89$,

We use the Congruent complementarity method to find that when A = 1, B = 3, which

$1^7\equiv 1\left(\mathrm{mod}17\right)$, $7\times 3^2\equiv 12\left(\mathrm{mod}17\right)$, $89\equiv 4\left(\mathrm{mod}17\right)$,

then $1+12+4\equiv 0\left(\mathrm{mod}17\right)$ and $w\equiv 0\left(\mathrm{mod}17\right)$ Finally, find $C=-w/17=-153/17=-9$, which is the solution to equation (24) are A = 1, B = 3, C = −9.

Example 4.2. Try to solving equation (25)

$64x^5+48x^4-32x^3+68x^2-10x-13=0$ (25)

By the equation (25), we obtain

$64x^5+48x^4-32x^3+68x^2-10x=13$ (26)

According to the Fundamental Theorem of Arithmetic, if equation (26) has an integer solution, 13 must contain an $x$ factor; But 13 is a prime number, substituting $x$ into the equation does not make the equation hold. So the equation may have a solution or it could be a real number. After observation, the coefficients on the left side of the equation are all even, so there may be 2 hidden factors on the right side of the equation, that is, 13 = 2 × 13 × 0.5. Perhaps $x=0.5$, substituting it into equation (5) actually satisfies the requirements of the equation.

$64\times {0.5}^5+48\times {0.5}^4-32\times {0.5}^3+68\times {0.5}^2-10\times 0.5=2+3-4+17-5=13$.

Obviously, when x = 0.5, both sides of equation (26) are equal, so x = 0.5 is a rational of the solution to equation (25).

Example 4.3. Try to solving equation (27)

$5A^7+7B^5+17C^3-5D^2+93=0$ (27)

According to the principle of L-algorithm,

$w=5A^7+7B^5+3C^3+93$, (28)

and it is best to consider D² = 25 and w = 125, then we obtain

$w=5A^7+7B^5+3C^3+93=125$,

Substituting A = 1, B = 1, C = 2 into equation (28) yields

$w=5A^7+7B^5+3C^3+93=5\times 1^7+7\times 1^5+3\times 2^3+93=125$,

$w-5D^2=0$,

$5D^2=w$,

$D^2=w/5=125/5$,

$D^2=25$,

$D^2=\sqrt{25}=5$.

a) In the first algorithm, for a univariate high-order equation, if x is not a known integer term factor, the equation may have no real number solution.

b) When the power exponent on the left side of the equation is congruent with the power exponent of a term on the right side, it cannot be used the index method.

c) In the equation (20), sometimes V can be the power of the lowest exponent.