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Questions tagged [square-numbers]

This tag is for questions involving square numbers. A non-negative integer $n$ is called a square number if $n = k^2$ for some integer $k$. Consider using with the [elementary-number-theory] or [number-theory] tags.

A number $n$ is a square number if and only if it is the square of an integer. That is, if $n = k^2$ for some integer $k$.

The name square number, or perfect square, comes from the fact that these particular numbers of objects can be arranged to fill a perfect square.

The square numbers begin with: $$0, 1, 4, 9, 16, 25, 36, 49, \ldots$$

The $k$th square number is given by $k^2$ with the zeroth square being $0$. Square numbers are strictly non-negative as $k^2 \ge 0$ for all integer $k$. There are $\lfloor \sqrt{n} \rfloor + 1$ square numbers in the range $[0, n]$.

References:

https://en.wikipedia.org/wiki/Square_number

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Find all functions $f$ such that if $a+b$ is a square, then $f(a)+f(b)$ is a square

Question: For any $a,b\in \mathbb{N}^{+}$, if $a+b$ is a square number, then $f(a)+f(b)$ is also a square number. Find all such functions. My try: It is clear that the function $$f(x)=x$$ satisfies the given conditions, since: …
math110
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What is the algebraic intuition behind Vieta jumping in IMO1988 Problem 6?

Problem 6 of the 1988 International Mathematical Olympiad notoriously asked: Let $a$ and $b$ be positive integers and $k=\frac{a^2+b^2}{1+ab}$. Show that if $k$ is an integer then $k$ is a perfect square. The usual way to show this involves a…
kdog
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Alternative proof that $\frac{a^2+b^2}{ab+1}$ is a square when it's an integer.

Let $a,b$ be positive integers. When $$k = \frac{a^2 + b^2}{ab+1}$$ is an integer, it is a square. Proof 1: (Ngô Bảo Châu): Rearrange to get $a^2-akb+b^2-k=0$, as a quadratic in $a$ this has two values: $a$ and $kb - a = (b^2-k)/a$. (The second…
quanta
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Numbers $n$ such that the digit sum of $n^2$ is a square

Let $S(n)$ be the digit sum of $n\in\mathbb N$ in the decimal system. About a month ago, a friend of mine taught me the…
mathlove
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Let $k$ be a natural number . Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.

Let $k$ be a natural number. Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers. I tried to prove this by supposing one of them is a square number and by substituting the corresponding $k$ value. But I failed to prove it. If we ignore…
Angelo Mark
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$n!+1$ being a perfect square

One observes that \begin{equation*} 4!+1 =25=5^{2},~5!+1=121=11^{2} \end{equation*} is a perfect square. Similarly for $n=7$ also we see that $n!+1$ is a perfect square. So one can ask the truth of this question: Is $n!+1$ a perfect square for…
anonymous
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Can any number of squares sum to a square?

Suppose $$a^2 = \sum_{i=1}^k b_i^2$$ where $a, b_i \in \mathbb{Z}$, $a>0, b_i > 0$ (and $b_i$ are not necessarily distinct). Can any positive integer be the value of $k$? The reason I am interested in this: in a irreptile tiling where the…
Herman Tulleken
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Can a number be equal to the sum of the squares of its prime divisors?

If $$n=p_1^{a_1}\cdots p_k^{a_k},$$ then define $$f(n):=p_1^2+\cdots+p_k^2$$ So, $f(n)$ is the sum of the squares of the prime divisors of $n$. For which natural numbers $n\ge 2$ do we have $f(n)=n$ ? It is clear that $f(n)=n$ is true for the…
Peter
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Is $100$ the only square number of the form $a^b+b^a$?

Conjecture: $100$ is the only square number of the form $a^b+b^a$ for integers $b>a>1$. In other words, $(a,b)=(2,6)$ is the only solution. Can we prove/disprove this? Observations: The solution mentioned should not come as a surprise, since…
TheSimpliFire
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Prove that none of $\{11, 111, 1111,\dots \}$ is the perfect square of an integer

Please help me with solving this : prove that none of $\{11, 111, 1111 \ldots \}$ is the square of any $x\in\mathbb{Z}$ (that is, there is no $x\in\mathbb{Z}$ such that $x^2\in\{11, 111, 1111, \ldots\}$).
Pechenka
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Math olympiad 1988 problem 6: canonical solution (2) without Vieta jumping

There is a recent question about this famous problem from 1988 on this forum, but I'm unable to respond to this because the subject is closed for me (insufficient reputation). Therefore this new post on the subject. here's the link to the earlier…
Rutger Moody
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Can $n!$ be a perfect square when $n$ is an integer greater than $1$?

Can $n!$ be a perfect square when $n$ is an integer greater than $1$? Clearly, when $n$ is prime, $n!$ is not a perfect square because the exponent of $n$ in $n!$ is $1$. The same goes when $n-1$ is prime, by considering the exponent of…
Paulo Argolo
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Is difference of two consecutive sums of consecutive integers (of the same length) always square?

I am an amateur who has been pondering the following question. If there is a name for this or more information about anyone who has postulated this before, I would be interested about reading up on it. Thanks. If $x$ is the sum of $y$ integers, and…
Bender
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Why is there a pattern to the last digits of square numbers?

I was programming and I realized that the last digit of all the integer numbers squared end in $ 0, 1, 4, 5, 6,$ or $ 9 $. And in addition, the numbers that end in $ 1, 4, 9, 6 $ are repeated twice as many times as the numbers that end in $ 0, 5$ I…
ESCM
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Is it known that all primes can be expressed as a square number minus a prime number?

i.e. Conjecture is that for every prime p, there exists an integer n such that $=^2−$ where q is prime. e.g. $57593 = 240^2 - 7$ I assume it's either known / false / an entirely uninteresting result given the infinite number of squares and…
rich
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