Add 168 to a number to get the square of a positive integer

Solution: According to the question, let this number be mCause.

m+168=n^2...1)

m+100=p^2...2), where n and p are positive integers.

1)-(2), obtain: (m+168)-(m+100)=168-100=68=n2-p 2

i.e. 68 = n 2-p 2 = (n + p) * (n - p).

Because 68 can only be decomposed into 68 = 34 * 2 = 17 * 4 = 68 * 1, and n+p>n-p>0

So n+p=34,n-p=2; or n+p=17, n-p=4; or n+p=68, n-p=1

Solving the above three equations respectively results in making n and p positive integers only when n + p = 34 and n - p = 2, and the solution is: n = 18 and p = 16

So m = p 2-100 = 16 2-100 = 156

So this number is 156

Let this positive integer be a

100+a=(10+b)^2=100+b^2+20ba=b^2+20b=b(20+b)

168+a=(12+c)^2=144+c^2+24ca=c^2+24c-34=c(c+24)-24b(b+20)=c(c+24)-24

b(b+20)=c(c+20)+4c-24 so when 4c=24, there are positive integer solutions on both sides.

c=6,b=6

When b=6, a=6*26=156

So this positive integer is 156

I'll talk about the easy way.

Adding 168 to a number gives the square of a positive integer, and adding 100 also gives the square of a positive integer.

The square of the numbers ending in 1,2,3,4,5,6,7,8,9,0 respectively, the resulting number ends in 1,4,9,6,5,6,9,4,1.

This number, after the ending number +8 and +0, should be 1,4,9,6,5,6,9,4,1,0.

So the single digit of this number can only be 1,6

1》Let's assume that it is 6, +100 is 6, it ends with 4 or 6 after opening the square, +168 is 4, and it ends with 2 or 8 after opening the square, because (168-100) 20=, so the difference between the 2 numbers obtained cannot exceed 3, so this number +100, after opening the square, it must be 6, and +168 opening the square must be 8

Because 168-100 20=<4, the single digit after the square must be 1

Draws 16 and 18

In other cases, analyze it yourself.

You should be able to understand this method: (in fact, the kind of method you don't understand is a more formal solution) Solution: Observation:

The difference between the squares of two consecutive natural numbers from small to large is found to be a set of all odd numbers.

That is, 168 100 68, is the sum of several consecutive odd numbers.

The sum of odd numbers is odd, and the sum of even odd numbers is even.

So 68 is an even sum of odd numbers.

If 68 is the sum of 2 consecutive odd numbers, then 68 33 35 is 68 33 35 (x+168) (x+100) because 17 2-16 2=33, 18 2-17 2=35

That is, this number plus 168 is 18 squared, and this number plus 100 is 16 squared.

The number is 156

If 68 is the sum of 4 consecutive odd numbers, there is no solution.

So the solution to this problem is 156

If you don't understand something, please send me a message.

Solution: Because 1 2-0 2=1-0=1

Therefore, it can be found that the difference between the squares of two consecutive natural numbers from small to large is the set of all odd numbers, that is, 168 100 68 is the sum of several consecutive odd numbers and odd numbers, and the sum of even odd numbers is even, so 68 is the sum of even odd numbers.

If 68 is the sum of 2 consecutive odd numbers, then 68 33 35 is 68 33 35 (x+168) (x+100) because 17 2-16 2=33, 18 2-17 2=35

That is, this number plus 168 is 18 squared, and this number plus 100 is 16 squared to solve this number is 156

You don't understand this, what grade are you in this year, I'll help you see the solution method that meets you, the following methods are similar, in fact, your method is already very good, I'll explain the drop-down to you:

1: Add 168 to a certain number to get the square of a positive integer, you get a+168=b 22: Add 100 to get the square of a positive integer, you can get.

a+100=c^2

Then there is the solution of these two equations, and there is a condition that b and c are both positive integers.

a+168=b^2;a+100=c^2

b^2-168=c^2-100

b+c)(b-c)=68

b+c and b-c are the same odd and even.

b-c=2b+c＝34

b=18,c=16

So this number is 156

This number is 156

The single digit of all squares can only be: 1, 4, 5, 6, 9

If a positive integer is added to 100 and 168, the single digit of the resulting number must be one of 1, 4, 5, 6, and 9.

Only positive integers with a single digit of 1 or 6 conform to this basic rule, and then they just make it up to 156.

It's to list equations. As follows:

Let this positive integer be a, the first square number is m, and the second square number is n, then:

a+100=m^ ①

a+168=n^ ②

obtained: n -m =68 (n-m)(n+m)=68 (n-m)(n+m)=2*34 (n-m)(n+m)=4*17

Then there are: n-m=2 n+m=34 n=18 m=16

n-m=4 n+m=17 n=31 2 m=13 2 If it doesn't fit the topic, drop it.

n-m=34 n+m=2 n=18 m=-16 If it doesn't fit the topic, drop it.

n-m=17 n+m=4 n=31 2 m=-13 2 If it doesn't fit the topic, discard it.

n=18 m=16

Put n=18 m=16 generationsGet a = 156

This positive integer is 156

156 Solution: Let a+100=x 2, a+168=y 2, so y 2-x 2=68, so (y-x)(y+x)=68 because x, y is an integer, so x=16 y=18, so a=156

I'll do it too.

Let adding 100 become a perfectly squared number of x 2;Adding 168 becomes another perfect square number of y 2.

y^2-x^2=168-100=68

y+x)(y-x)=68

y+x)(y-x)=68*1 (y+x)(y-x)=34*2 (y+x)(y-x)=17*4

y+x=68, y-x=1 y+x=34, y-x=2 y+x=17, y-x=4

The solution of this system of equations is non-integer and does not fit the topic; y=18,x=16 The solution of this system of equations is non-integer and is not in place.

According to the title, 18 2-168 = 324-168 = 156

Therefore, this positive integer should be 156. Right, no.

Solution: Unknowns are constantly added in the process of solution, please look carefully at the process 100 10 2

Let this number be m, then.

100 m 10 2+20n+n 2, i.e.: m 20n n 2;

168+m=10^2+68+m=10^2+68+n^2+20n=10^2+20x+x^2

68+n^2+20n=20x+x^2

x^2-n^2+20(x-n)=68

x-n)*(x+n)+20(x-n)=68(x-n)(x+n+20)=68

Since m is a positive integer, x must be greater than n, so 68 can be divided into the product of those numbers? 1×68，2×34，4*17?

Obviously, (x n 20) is greater than 20, therefore, 4 17 and later numbers are excluded, so there are only the first two numbers, the solution x-n = 1 x + n + 20 = 68 as well.

x-n=2 x+n+20=34

The last solution: the first solution x is not a positive integer, excluded).

The second system of equations: x 8, n 6, brings in the top m 20n n n 2 = 156

So, the answer is 156

Solution: Let this positive number be a, then a+100=x 2, a+168=y 2y2-x2=68

y-x)(y+x)=68

Because if x, y is an odd and an even, then y+x, y-x are all odd numbers, and 68 is an even number, so y+x and y-x should be the same odd or even, and the 68 factorization is decomposed because it is the same odd and even, so only the decomposition is 2 and 34 so: y+x=34, y-x=2 (y+x is greater than y-x) y=18

So from a+168=y 2:

a=156A: This positive number is 156.

Let the arithmetic square roots of the two perfect squares obtained be a and b(a < b) respectively, then according to the meaning of the question:

a 2 - 100 = b 2 - 168 deformation :

b^2 - a^2

b + a)(b - a) = 168 - 100 = 68 = 2×2×17

Because b + a and b - a must have the same parity, according to the factor of 68, we know that b + a and b - a must be dual.

And because b + a > b - a get:

b + a = 2×17

b - a = 2

This is the problem that requires a binary equation).

Solution: b = 18

a = 16

So this positive integer = 16 2 - 100 = 18 2 - 168 = 156

Solution: Let this positive number be a, then a+100=x 2, a+168=y 2y2-x2=68

y-x)(y+x)=68

Because if x, y is an odd and an even, then y+x, y-x are all odd numbers, and 68 is an even number, so y+x and y-x should be the same odd or even, and the 68 factorization is decomposed because it is the same odd and even, so only the decomposition is 2 and 34 so: y+x=34, y-x=2 (y+x is greater than y-x) y=18

So from a+168=y 2:

a=156A: This positive number is 156.

156, these two numbers are 16 and 18, just guess the number.

Let this positive integer be a, according to the meaning of the title 100+a=b 2

168+a=c^2

So c 2-b 2 = 68

c+b)(c-b)=2×2×17

So there are three possibilities: c-b=1, c+b=68c-b=2, c+b=34

c-b=4 c+b=17

One or three of the two may result in b and c as scores, so they are not in line with the topic and are discarded.

From c-b=2 c+b=34, c=18, b=16, therefore, a=b, 2-100=156

156 square number divided by 4 is either divisible or remainder 1 plus 100 at a time plus 168 all cross at least one square number rounding the first square number is a second time if it is (a+2) then 168-100 = (a + 2) -a =4a + 4 solution a = 16 16 -100 = 156 if it crosses 3 squares 168-100 = (a + 4) -a = 8a + 16 a is not an integer so it can only be 156

Haha, it's 156, you check it, right?

Let this number be n, and n+168=a2, n+100=b2, then a2-b2=68=22 17, i.e. (a+b)(a-b)=22 17 But a+b and a-b have the same parity, so a+b=34, a-b=2, so a=18, b=16, and thus n=156

So the answer is 156

The squared number above 100 is 121,144,169,196,225,256,289,324,361,400 ,......

100 + 21 = 121 is, 168 + 21 = 189 is not;

100 + 44 = 144 yes, 168 + 44 = 212 is not;

100 + 69 = 169 is, 168 + 69 = 237 is not;

100 + 96 = 196 is, 168 + 96 = 264 is not;

100 + 125 = 225 yes, 168 + 125 is not;

100 + 156 = 256 is, and 168 + 156 = 324 is also.

This positive integer is 156