1, it is known that n is a positive integer, 2n+ 1 and 3n+ 1 are completely squared, so:

n=40，

5n+3=5*40+3=203

Because 203=29*7 is not a prime number.

So there is no such number n; ? ##

2. Let m be an integer, and the equation MX 2+2 (m-5) x+m-4 = 0 about x has integer roots, so what is the value of m? (m=- 18)

= = & gtdelta:=[2(m-5)]^2-4m(m-4)= 100-24m

The solution of the original formula: x = [-2 (m-5) √ (100-24m)]/2m.

=- 1+[5√(25-6m)]/ m

=- 1+{5 √[5^2+(-6m)]}/m？

To make √ [5 2+(-6m)]} an integer,

= =>5 2+(-6m) must be a complete square number.

= => From Pythagoras number 5- 12- 13, we get

-6m= 12^2= 144

m =- 18；

= = & gt？ x =- 1+{ 5 √[5^2+(-6*- 18)]}/(- 18)

=- 1+{5? √[5^2+ 12^]}/(- 18)

=- 1+(5 ? 13)/(- 18)

There is an integer root: =-1+(5+13)/(-18) =-2;

3. If for a natural number n not less than 8, when 3n+ 1 is a complete square number, n+ 1 can be expressed as the sum of k complete squares, then the minimum value of k is (? 1? )

a、 1？ b、2C、3D、4

Solution:

When 3n+ 1 is a complete square number,? N+ 1 can be expressed as the sum of k complete squares,

A natural number n not less than 8, where n=8, with:

3*8+ 1=25 is a complete square number;

n+ 1 = 8+ 1 = 9；

9=3^2=2^2+2^+ 1^2;

So the smallest k =1; ?

4. If m 2 = n+2 and n 2 = m+2 (m is not equal to n), the value of m 3-2mn+n 3 is (? 0? )

a、 1B、0C、 1D、2

Solution: m 2 = n+2, n 2 = m+2, two expressions are subtracted: (m2-N2) =-(m-n) = > m+n =-1;

M 2 = n+2, n 2 = m+2, and the two expressions are added: (m2+N2) = (m+n)+4 = = > m 2+n 2 = 3;

Because: m+n =-1= = > (m+n)^2=(- 1)^2

= = & gt？ m^2+n^2+2mn= 1

= = & gt？ mn=[ 1-(m^2+n^2)]/2=( 1-3)/2=- 1；

m+n =- 1 = = & gt； (m+n)^3=(- 1)^3

= = & gt？ m^3+n^3+3mn(m+n)=- 1

= = & gt？ m^3+n^3= 1-3mn(m+n)= 1-3*(- 1)(- 1)=-2；

So: m3-2mn+n3 =-2-2 * (1) = 0; ? ##

5. Let N=23x+92y be a complete square number, and n does not exceed 2392, then all positive integer pairs (x, y)*** have _2 1 15_ pairs that satisfy the above conditions.

Solution: Because N=23x+92y,?

= = & gty=-x/4+N/92

Because n does not exceed 2392

So n/92 < = 2392/92 = 26;

In contrast, the possible range of N/92 (26, 25, 24, 23, 22…, 3, 2, 1) is: only when N/92=23, N/92=23.

= =>n = 2116 = 46 * 46, which is a complete square number.

= = & gty=-X/4+2 1 16

That is, find the positive integer solution (x, y) on the straight line y=-X/4+2 1 16.

= => The general solution of its positive integer:? (X=4K, Y=2 1 16-K), where (k is a natural number, K= 1, 2,3,, n).

To make Y=2 1 16-k a positive integer,

= = & gty = 2 1 16-k & gt； 0;

= = & gtK & lt2 1 16; K=2 1 15？ ;

So * * * has 2 1 15 pairs of positive integers (x, y); ##

6. In the plane rectangular coordinate system xOy, we call the points whose abscissa is an integer and whose ordinate is a complete square number as "good points", and find the coordinates of all "good points" on the quadratic function Y = (x-90) 2-4907 image.

(Whether "4907" in the topic "Y = (X-90) 2-4907" is wrong, please check it carefully and modify it! ! ! )

7. It is known that the roots of the equation x 2-6x-4n 2-32n = 0 are all integers, so find the value of the integer n. ..

Solution: = => delta: = 6 2-4 (-4n 2-32n) = 36+4 (4n 2+32n).

Solution of the original formula: x = 6 √ [36+4 (4n 2+32n)]/2.

=3 √(4n^2+32n+9)

To make x an integer,

= = & gt4n 2+32n+9 must be a complete square number.

= = & gtGet: take 4n 2+32n+9 = (1,4,9,16,25,36,49,64, …, n 2).

4n^2+32n+9=9

= = & gtn = 0； ##

8. If D, E and F are points on BC, CA and AB of △ABC, respectively, BD: DC = 1, Ce: EA = 2, AF: FB = 3 and S△ABC=24, find the area of △DEF.

Solution:

(1) found S3.

△ABC, △AFC and △BFC have the same height with AB as the base and passing through point C, and are set as H.

So there is: AB*H=? s△ABC；

FB*H=？ s△BFC； ?

Divided by two formulas: S△BFC=FB/AB*? s△ABC；

Because of AF? :FB = 3； ? = = & gtAB:FB = 4；

So: S△BFC=FB/AB*? s△ABC = 1/4 * 24 = 6；

At △BFC, d is the midpoint of BC, so:

S3 and S△DFC have the same area, = =>? S3=？ s△BFC/2 = 6/2 = 3；

② Find S2, S 1?

△ABC, △ABE, and △BEC are based on AC and pass through point B, and their heights are the same, so they are set as Hb.

So there is: AC*Hb=? s△ABC； ? - (*)

AE*Hb=？ S△ Abe; ? - (**)

EC*Hb=？ s△BEC； ? - (***)

Divided by (*) and (* *): S△ABE=AE/AC*? s△ABC；

Divided by (*) and (* *): s △ BEC = CE/AC *? s△ABC；

Because CE:AE? =2; ? = = & gtAE:AC = 1/3；

= = & gtCE:AC = 2/3；

So: S△ABE=AE/AC*? s△ABC = 1/3 * 24 = 8；

S△BEC=CE/AC*？ s△ABC = 2/3 * 24 = 16；

In △ABE, f is the (3: 1) point of AB, so: (Similarly, the height is equal, but the bottom is different).

The ratio of S2 to S△DFC area = the ratio of bottom edge =AF/FB=3: 1.

= = & gt？ S2 and? The ratio of S△ABE = 3/4;

= = & gt？ S2=？ s△ABE * 3/4 = 8 * 3/4 = 6；

Similar S 1=? s△BEC * 1/2 = 16 * 1/2 = 8；

So s △ def = s △ ABC-s1-S2-S3 = 24-8-6-3 = 7; ##

9. Let a 2+1= 3a, b 2+1= 3b, a≠b, then the value of algebraic expression (1/a 2)+(1/b 2) is (. B=7？ )

a、5B、7C、9D 1 1

Solution: A 2+ 1 = 3a, B 2+ 1 = 3b subtraction.

= = & gta^2-b^2=3(a-b)

= = & gt(a-b)(a+b)=3(a-b)，？ And a≠b,

= = & gta+b=3( 1)

A 2+ 1 = 3a，B 2+ 1 = 3b。

= = & gta^2+b^2+2=3(a+b)

= = & gta^2+b^2=3*3-2=7； (2)

Because (1)a+b=3.

= = & gt(a+b)^2=3^2=9

= = & gta^2+b^2+2ab=9；

= = & gt？ 2ab=9-(？ a^2+b^2)=9-7=2；

= = & gt？ ab = 1； ; ?

So (1/a2)+(1/b2) = (a2+b2)/(? ab)^2=7/ 1=7； ? ##