All integers that make x 2 5x 24 perfectly squared are 50

Factoring, x 2-5x-24 = (x-8) (x+3), when x 2-5x-24 = 0, x = 8 or -3

When x 2-5x-24≠0 and (x-8, x+3) = (11, x+3) = 1 or 11.

When (x+3,x-8)=11, let x+3=11*a 2, x-8=11*b 2, then.

11=11*a2-11*b2, ie.

a+b)(a-b)=1, since a and b are integers, obtain.

a+b=1 and a-b=1, so a=1, b=0, x=8, rounded.

When (x+3,x-8)=1, let x+3=a2, x-8=b2, then.

11=a 2-b 2=(a-b)(a+b), so a+b=11, a-b=1, so.

a=6, b=5, so x=33.

The combined score is x and -3, and the number is 3

When solving this kind of problem, it is usually necessary to factor it, find the common divisor of each term, and then classify it as a 2 and so on, and solve the relationship between a 2 and b 2, so as to find x. This problem is special because there is an integer that has a value of 0 and needs to be classified, otherwise -3 will be missed.

When encountering this problem, it is actually a matter of splitting the item or making up the number.

x^2-5x-24

x^2-5x+（5/2）^2-（5/2）^2-24x-5/2)^2-121/4

x+3)(x-8)

x 2-5x-24=(x+3)(x-8)=[(,m is an integer, so (,so(2x-5) 2=(2m) 2+11 2,when m=0, x=8, or x=-3 ; When m≠0, (2x-5) 2=(2m) 2+[(6+5)(6-5)] 2=(2m) 2+(6 2-5 2) 2=4(m) 2+(6 2) 2+(5 2) 2-2(6 2)(5 2), so when m 2=(6 2)(5 2), (2x-5) 2=(6 2+5 2) 2, i.e., 2x-5= 61, i.e., x=33, or x=-28, To is all integer solutions for x, four in total.

0, you make y1=x 2-5x-24, y2=x 2,, on the image, find their intersection, then let y1=y2

We find that x= has only one intersection point, and this number is not an integer, so it is 0.

x 2-5x-24 becomes the number of full flat sheds to return to the royal square, x 2-5x-24>=0

1) x 2-5x-24=0, which gives x=8 or x=-3

2)x^2-5x-24>0

x^2-5x-24=(x-8)(x+3)

Let (x+3) = n 2, (x-8) = m 2, subtract n 2-m 2 = 11, i.e. (n-m)(n+m) = 11

When (n-m)=1, (n+m)=11, the solution is n=6, m=5, corresponding to x=33

n-m)=11, (n+m)=1, the solution is n=6, m=-5, corresponding to x=33

n-m)=-1, (n+m)=-11, the solution is n=-6, m=-5, corresponding to x=33

n-m)=11, (n+m)=1, the solution is n=6, m=-5, corresponding to x=33

Require(x+3)=-n2,(x-8)=-m2

Subtract m2-n2=11, i.e., (m+n)(m-n)=11

When (m+n)=1, (m-n)=11, the solution is m=6, n=-5, corresponding to x=-28

m+n)=11, (m-n)=1, the solution is m=6, n=5, corresponding to x=-28

m+n)=-1, (m-n)=-11, the solution is m=-6, n=5, corresponding to x=-28

m+n)=11, (m-n)=1, the solution is m=6, n=5, corresponding to x=-28

Total x = 8, -3, 33, -28

The value of the square of fraction 4x-20 x -2x-15 is an integer when the integer x is.

The calculation process is as follows: 4x-20 x 12x-15 to make 20 x an integer, and x is also an integer, only when x= 1 or perturbation 2, the value of the fraction is an integer. Suspension of the book.

Original = [(x+1)(x+4)][x+2)(x+3)]+1[(x +5x)+4][(x +5x)+6]+1(x +5x) +10(x +5x)+24+1(x +5x) +10(x +5x)+25(x +5x+5).

So it's a perfectly squared number of an integer.

x+1)(x+2)(x+3)(x+4)+1(x²+5x+4)(x²+5x+6)+1

x²+5x)²+10(x²+5x)24+1(x²+5x)²10(x²+5x)+25

x²+5x+5)²

When x is an integer, (x+1)(x+2)(x+3)(x+4)+1 is a perfectly squared number of an integer.

x+1)(x+4)=x2+5x+4

x+2)(x+3)=x2+5x+6

Let x2+5x be t, then the original algebraic formula is (t+4)(t+6)+1, and the decomposed t2+10t+25, i.e., (t+5)2

Since x is an integer, t+5 is an integer, i.e., it is proven.

x 2-5x-24 becomes a perfectly squared number, x 2-5x-24 > = 0

1) x 2-5x-24=0, which gives x=8 or x=-3

2)x^2-5x-24>0

x^2-5x-24=(x-8)(x+3)

Let (x+3) = n 2, (x-8) = m 2, subtract n 2-m 2 = 11, i.e. (n-m)(n+m) = 11

When (n-m)=1, (n+m)=11, the solution is n=6, m=5, corresponding to x=33

n-m)=11, (n+m)=1, the solution is n=6, m=-5, corresponding to x=33

n-m)=-1, (n+m)=-11, the solution is n=-6, m=-5, corresponding to x=33

n-m)=11, (n+m)=1, the solution is n=6, m=-5, corresponding to x=33

Require(x+3)=-n2,(x-8)=-m2

Subtract m2-n2=11, i.e., (m+n)(m-n)=11

When (m+n)=1, (m-n)=11, the solution is m=6, n=-5, corresponding to x=-28

m+n)=11, (m-n)=1, the solution is m=6, n=5, corresponding to x=-28

m+n)=-1, (m-n)=-11, the solution is m=-6, n=5, corresponding to x=-28

m+n)=11, (m-n)=1, the solution is m=6, n=5, corresponding to x=-28

Total x = 8, -3, 33, -28

Finishing: (x+1)*(x+2)*(x+3)*(x+4)+1=[(x+1)(x+4)][x+2)(x+3)]+1=(x 2+5x+4)(x 2+5x+6)+1=(x 2+5x) 2+10(x 2+5x)+24+1=(x 2+5x) 2+10(x 2+5x)+25=(x 2+5x+5) 2

Since x is an integer, x 2+5x+5 is also an integer, so it is a perfectly squared number of an integer.

Solution:

（x+1)(x+2)(x+3)(x+4)+1

=[(x+1)(x+4)][x+2)(x+3)]+1

x²+5x+4)(x²+5x+6)+1=(x²+5x)²+10(x²+5x)+24+1=(x²+5x)²+10(x²+5x)+5²=(x²+5x+5)²When x is an integer, x +5x+5 is an integerSo (x+1)(x+2)(x+3)(x+4)+1 is a perfect square number of an integer

(x+1)(x+2)(x+3)(x+4)+1=[(x+1)(x+4)][x+2)(x+3)]+1=(x 2+5x+4)(x 2+5x+6)+1 reg x 2+5x=t

t+4)(t+6)+1

t^2+10t+25

t+5)^2

x^2+5x+5)^2

So the integer (x+1), (x+2), (x+3), (x+4)+1 must be perfectly squared.

Let 25x 2-1320x+20=r 2, r be an integer. (5x-132) 2 = 25x 2-1320x+20+17404. Let y=5x-132, and y is also an integer.

then, y2 = r2 + 17404, gets: (y+r)(y-r)=17404. Decompose 17404 to get 17404=229*19*2*2*1, so there is:

a): y+r)(y-r)=17404*1=(-17404)*(1)

or (b): y+r)(y-r)=8702*2=(-8702)*(2).

or (c): y+r)(y-r)=4351*4=(-4351)*(4).

or (d): y+r)(y-r)=916*19=(-916)*(19).

or (e): y+r)(y-r)=458*38=(-458)*(38).

or (f): y+r)(y-r)=229*76=(-229)*(76).

From these formulas, we can see that y+r=a, y-r=b, where a, b are the numbers decomposed above, so y=(a+b) 2. Since y is an integer and a+b must be even, only the formulas (b) and (e) are possible in the above equation.

From equation (b), 5x-132=y=+ or -(8702+2) 2=+ or -4352, an integer solution is solved: x=-844.

From equation (e), 5x-132=y=+ or -(458+38) 2=+ or -248, an integer solution is solved: x=76.

There are only these two solutions.

The above solution is for reference.