All Questions
12
questions
2
votes
3
answers
177
views
Is $1 \times 1 + 2 \times 2 + 3 \times 4 + 4 \times 8 = 49$ a coincidence? (Is $\sum_{i=0}^k(i+1)2^i$ ever a square again?)
When watching a gaming video, I noticed an intriguing fact:
$$
1 \times 1 + 2 \times 2 + 3 \times 4 + 4 \times 8 = 49,
$$
which is a square number.
I asked myself, is this a coincidence or not? ...
- 4,051
5
votes
0
answers
73
views
Is $1105$ the only Poulet-number of the form $2^a+3^b$?
Conjecture : $1105$ is the only Poulet-number that can be written in the form $2^a+3^b$ with positive integers $a,b$
A Poulet-number is a composite number $N$ satisfying the congruence $2^{N-1}\equiv ...
- 85.1k
1
vote
0
answers
408
views
Sum of perfect powers of n natural numbers
The perfect power of an integer $n = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}$ are $p_1^2, p_1^3\cdots p_1^{\alpha_1},p_2^2, p_2^3 \cdots p_2^{\alpha_2},(p_1 p_2)^2, (p_2 p_3)^2\cdots$ ...
- 49
8
votes
1
answer
265
views
Surprising fact about a certain number-theoretic function
Ante suggested the following function :
For natural number $n$ we can observe the $n$ remainders $b_1,...,b_n$ by writing $n$ as $n=a_k \cdot k+b_k$ for $1 \leq k \leq n$
Because of the familiar ...
- 85.1k
2
votes
0
answers
78
views
Are there forced factors of numbers of this kind?
Let $$f(n):=n^{n^2}+(n+1)^{(n+1)^2}$$ for a positive integer $n$. For clarification $n^{n^2}$ means $n^{(n^2)}$ , analogue for the other summand.
Can we find a concrete factor of $\ f(n)\ $ (like ...
- 85.1k
3
votes
0
answers
132
views
Are there infinite many pairs of primes $(p/q)$ such that $p^3+q^3$ is a perfect square?
The pairs of primes $(p,q)$ with $p\le q\le 10^5$ such that $p^3+q^3$ is a perfect square , are (the last entry is the maximum possible exponent of the perfect power $p^3+q^3$ which is only different ...
- 85.1k
8
votes
0
answers
187
views
Is $2^9$ the only power of two that is the sum of two odd perfect powers?
Let $m\ge 1$ be an integer.
For which $m$ can we find odd perfect powers $a,b$ with $a+b=2^m$ ? The only solution, I found for $m\le 80$ is $m=9$ with the representation $$2^9=13^2+7^3$$
The ...
- 85.1k
12
votes
0
answers
166
views
Can a prime have arbitary many representations as a sum of two perfect powers?
Let $p$ be a prime number and $f(p)$ be the number of representations $$p=a+b$$ with perfect powers $0<a< b$
For example, $13$ has the only representation $$4+9=13$$ hence $f(13)=1$
$$41=9+32=...
- 85.1k
2
votes
0
answers
64
views
Which cycles are possible by repeated summation of the cubes of the digits of a number?
Here :
Digital root with squared digits
the possible cycles of repeated summation of the squares of the digits of a number are mentioned. What about cubes ?
$1$ , $153$ , $370$ , $371$ and $407$ ...
- 85.1k
12
votes
5
answers
301
views
Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$?
Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase ...
- 13.2k
7
votes
2
answers
1k
views
proof - Show that $1! +2! +3!+\cdots+n!$ is a perfect power if and only if $n=3$
Show that $1! +2! +3!+\cdots+n!$ is a perfect power if and only if $n =3$
For $n=3$, $1!+2!+3!=9=3^2$. I also feel that the word 'power' makes it a whole lot hard to prove. How do we prove this? What ...
- 4,032
5
votes
0
answers
522
views
Sum of like powers equal to a power
It's not hard to prove that $$(1+2+3+\ldots+n)^2=1^3+2^3+\ldots+n^3$$ ( for example using induction )
A generalization of this is also known :
$$(\sum_{d \mid n} \tau(d))^2=\sum_{d \mid n} \tau(d)^3$...
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