The answer to my question is only 1!
Is there anything other than 1?
The answer is NO! how to prove? Notice that squares of integers are of the form 4k or 4k+1 only, whereas 11, 111, 1111,... They're all in the form 4k+3, so they're not absolute squares.
What are absolute squares?
Prove that a^(4b+1)-a is always divisible by 30 foor whatever a and b
I will prove that a^(4b+1) divisible by 2, 3 and 5.
If a is even, a^(4b+1) is also even. So a^(4b+1)-a divisble by 2.
If a is odd, a^(4b+1) is also odd. So a^(4b+1)-a divisble by 2.
Therefore 2 | a^(4b+1)-a for all a,b is posite integer.
What are absolute squares?
Prove that a^(4b+1)-a is always divisible by 30 foor whatever a and b
I will prove that a^(4b+1) divisible by 2, 3 and 5.
If a is even, a^(4b+1) is also even. So a^(4b+1)-a divisble by 2.
If a is odd, a^(4b+1) is also odd. So a^(4b+1)-a divisble by 2.
Therefore 2 | a^(4b+1)-a for all a,b is posite integer.
Proving that 3 | a^(4b+1)-a
a^(4b+1)-a = a(a^4b)-a = a(a^4b -1) = a(a^2b -1)(a^2b +1)
case 1: 3 | a ; we have 3 | a(a^2b -1)(a^2b +1)
case 2: a is not divisible by 3 ; then a^2b not divisible by 3 too. So a^2b -1 or a^2b +1 either of them must be divisible by 3 that make 3 | a(a^2b -1)(a^2b +1).
Proving that 5 | a^(4b+1)-a
a^(4b+1)-a = a(a^4b)-a = a(a^4b -1)
case 1: 5 | a ; clearly 5 | a(a^4b -1)
case 2: a ≡ 1,2,3 or 4 (mod 5) ; will get a^4 ≡ 1 (mod 5). So a^4b -1 ≡ 1^b -1 ≡ 0 (mod 5) then 5 | (a^4b -1) and 5 | a(a^4b -1) according to the problem.
So we can conclude that 30 | a^(4b+1)-a for all a,b ∈ N.
It's been for 3 months that I come back to prove it successfully. 
Prove that if 3|4x-y implies that 9|4x² + 7xy - 2y²
P.S. I did this by modular arithmetic , I.e. by proving all the cases
I am not disclosing the method I used so as not to disrupt the mathematical experience you will go through by solving this problem
THIS IS INTERSTING PROBLEM!
Consider 4x² + 7xy - 2y² = (4x-y)(x+2y) = (4x-y)(4x-y-3(x-y))
From 3 | 4x-y , there is an integer a such that 3a = 4x-y.
So (4x-y)(4x-y-3(x-y)) = 3a(3a-3(x-y)) = 3a(3(a-x+y)) = 9a(a-x+y) is divisible by 9.
If 4 whole numbers taken at random are multiplied what is the probability that the last digit of the product is 1 or 3 or 7 or 9
It's not difficult to see that all 4 random numbers are the unit digit must be 1,3,7 or 9 only. Consider choosing the unit digit of the random number.
Event -> n(E) = 4⁴ = 256
Sample space -> n(S) = 10⁴ = 10,000
So posibility is n(E)/n(S) = 256/10,000 = 0.0256
Problem 5:
How many integers n from 1 to 100 that make the equation n³+n² = x² have a solution x in integer?
9, if you mean 1 to 100 inclusive
Rearranging gives us
n*sqrt(n+1)=x
Only the numbers one less than a perfect square will give us an integer solution. n=1 does not work because it will give us a square root of 2. Fun problem (hopefully I got it right
Rearranging gives us
n*sqrt(n+1)=x
Only the numbers one less than a perfect square will give us an integer solution. n=1 does not work because it will give us a square root of 2. Fun problem (hopefully I got it right
That correct!
Problem 6:
Give four examples of prime factors of 2⁶⁴-1.
- or +?
Only positive.
Problem 7:
Find the smallest positive integer number that multipled by 8! (8 factorial) Then it's a perfect square number.
Ummm. So that would mean x•8! = a•a
8! = 8×7×6×5×4×3×2×1 = (4×4)(2×2)(3×3)(2)(7×5)
= 4² • 2² • 3² •7•5•2
We have to find the smallest multiple of 7•5•2 = 70 which is a square number.
Now since these 3 are prime numbers, we can't multiply say ,2, with another natural number to get 5. So the smallest suchnumber is 70 itself.
8! • 70 = 1680²
That right!
SOLUTION: problem 6
Give four examples of prime factors of 2⁶⁴-1.
2⁶⁴-1
= (2³²-1)(2³²+1)
= (2¹⁶-1)(2¹⁶+1)(2³²+1)
= (2⁸-1)(2⁸+1)(2¹⁶+1)(2³²+1)
= (2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)
= (15)(17)(257)(2¹⁶+1)(2³²+1)
Notice that 3, 5, 17, 257 are all prime factors of 2⁶⁴-1.
Problem 8:
Prove that the product of two adjacent postive integers is not a perfect square.
Notation- I will use the term "number" interchangeably with "positive integer" as is the convention in number theory.
Proposition 1: Let a² be a square number.
And then in the prime factorization of a², all the primes p, q, r, ... etc. Will be raised to even powers.
Now,
Prop.2:
adjacent numbers cannot have a common divisor (except for 1) .
Prop.3:
Their is only one prime factorization for numbers.
Prop.4:
If {p, q, r, ...} is the set of all the prime factors of n, then all the divisors of n are multiples of different degrees of p, q, r ....
Prop.4 Follows from prop.3 ,
Suppose A is a divisor of n. The n, all the primes that divide A divide n ( if a divides b, then all the divisors of a divide b too) and the only primes that divide n are the ones in its prime factorization (there is only one prime factorization of a number). There fore A is a product of p, q, r etc.
Now let a and a+1 be the adjacent numbers.
And (p,q,r,...) be the prime factors of a and (s,t,u..). be the prime factorization of a+1. Then a+1•a = (p,q,r...) × ( s,t,u...,) and since these are all different primes, their product cannot be a square number.
Method2: a×a+1 = a²+a (by the distributive property). So the product of two adjacent numbers will always be greater than a square number by the smaller number.
Absolutely clear!
This is my way.
Let n be any positive integer. The product of two adjacent numbers has the form n(n+1).
Notice that n² < n(n+1)=n²+n < (n+1)²
Obviously, there is no perfect square number between n² and (n+1)².
So n(n+1) is not perfect square for all n ∈ Z+
Problem 9:
Prove that if n and n²+2 are prime numbers, then n³+4 is also a prime number.
What are absolute squares?
Prove that a^(4b+1)-a is always divisible by 30 foor whatever a and b
I will prove that a^(4b+1) divisible by 2, 3 and 5.