Is a number meeting these conditions divisible by forty-nine? I am not a mathematician, I'm a linguistics PhD student. As part of my research I need to put various convoluted sentences through various syntactic transformations and see then check whether people think they are true or not. Mathematical statements (well, some of them) suit my purposes very well, because they are less context dependent and can be straightforwardly assigned a truth value (i.e. be deemed true or false). The problem is that I'm not a mathematician. When these sentences get a bit convoluted, I have a bit of a problem knowing whether they are true or false myself (before they undergo various syntactic transformations). 
I have a particular sentence which states that if a given number is:


*

*an integer

*divisible by 7 (meaning it will yield an integer if divided by 7)

*a square number


then it is divisible by 49. I intuitively believe this to be correct (although I can't explain why). Is this actually true? I don't want to waste everybody's time by starting with an untrue untransformed sentence.
 A: Yes. The reason is that the only way a square number can be divisible by $7$ is if its square root is divisible by $7$. So your number is the result of squaring a multiple of $7$, and when you do that you get a multiple of $49$.
A: You can derive this fairly directly from Euclid's Lemma, which says that if a product $a \cdot b$ is divisible by prime $p$, then either $a$ or $b$ (or both) is divisible by $p$.
So, $n^2$ divisble by 7 means $n \cdot n$ divisible by 7 which (by Euclid's Lemma) means $n$ divisible by 7 and thus $n^2$ divisible by 49.
Of course it took even Euclid a little work to prove the Lemma
A: Yes.
To see this, call the number $x$ and suppose that $x$ is the square of the number $y$. Suppose that $y$ has a prime factorization
$$y = p_1^{n_1}p_2^{n_2}\cdots p_K^{n_K}$$
where each $n_k>0$ and $p_1<p_2<\cdots < p_K$. (In other words, we're displaying the prime factorization as compactly as possible, and "in order".)
Since $x = y^2$, we have
$$x = (p_1^{n_1}p_2^{n_2}\cdots p_K^{n_K})^2 = p_1^{2n_1}p_2^{2n_2}\cdots p_K^{2n_K}$$
Note all of the powers $2n_1,2n_2,\ldots,2n_K$ are even; and since $7$ divides $x$, one of the prime factors $p_i$ must be $7$.
But then $2n_i$ is an even number greater than zero, so is at least $2$. Thus there are at least two factors of $7$ in $x$, so $49$ divides $x$.
A: Yes, it is true. It hinges on the fact that $7$ is a prime number.
In general, if $n$ is an integer that is divisible by a prime number $p$ and $n$ is a square, then $n$ is divisible by $p^2$.
This follows from the 
Fundamental theorem of arithmetic.
A: This is true. Let us call your number $n$. The number $n$ is supposed to be the square of some number, say $a$, so $n=a^2$. Now since $a^2$ is divisible by $7$ and $7$ is prime, $a$ has to be divisible by $7$. (It is always true that if a prime number divides a product, it has to divide one of the factors.) But if $a$ is divisible by $7$, then $a^2$ is divisible by $7^2=49$.
A: Definition: A square number is a number whose square root is a whole number.
Definition: The square root of a number, call it N, is a number which can be created by squaring some other number, call it m, i.e., if m^2 = N then sqrt(N) = m 
Fundamental Theorem of Arithmetic (FTA): Any whole number greater than 1 (i.e., from 2 up) can be written as a unique product of prime numbers (the unique part means no matter how you split it up you get the same list in the end).
So, if you start with a (perfect) square whole number, call it N, being divisible by 7:
Since N is a square, it must be factorable as a product of some number times itself, i.e., N = m*m. Now, each m is either prime or factors again into a product of primes (by FTA). Once you have this list of primes written out, 7 must be there in that list as least once, since 7 divides N. But, since both m's are the same m, the 7 must appear twice (once for each m), i.e., since there's a pair of m's there must be a pair of all the factors of each m. 
This is just a wordy version of what user MPW was saying using formulas.
Also, this can be generalized for any prime number, not just the number 7.
Now that I think about it, it could be generalized to any whole number, not just primes.
Also, a caveat, your original post says the number should be an integer. If the original number is negative, it cannot be a perfect square (unless you want to talk about complex numbers). I suspect you meant "whole number", i.e., just positives.
A: One simple way of expressing the fundamental theorem of arithmetic is this.
Let $p_i$ be the ith prime number. Then every positive integer, $n$, can be expressed uniquely as an infinite product
$$n = \prod_{i=1}^\infty p_i^{\pi_i}$$
where all of the $\alpha_i$ are non negative integers and all but a finite number of them are equal to $0$. (A fancy way of saying this is that the sequence $\{\pi_i\}_{i=1}^\infty$ has finite support.
Suppose $n$ is:


*

*an integer

*divisible by 7, which equals $p_4$.

*a square number


Then there must be an integer $m$ such that $n = m^2$
We know that, for some $\alpha_i$, 
$$m = \prod_{i=1}^\infty p_i^{\alpha_i}$$
in which case,
$$n = \prod_{i=1}^\infty p_i^{2\alpha_i}$$
Since $n$ is a multiple of $7$, then $2\alpha_4 > 0$. Hence $\alpha_4 > 0$
Since there is the only way to express $m$, then $m$ is a multiple of $7$.
A: suppose this number is x. 
Now we know:


*

*x is an integer (Assumption: positive integer)

*x is divisible by 7

*x is a square number





*

*From Point 1 we know its a positive integer. From that we can say x is 1 or greater than 1. (x=1 or x>1)

*From Point 2 we know x is divisible by 7. That brings us to conclude that x is at least 7. Or multiple of 7 since its divisible by 7. We can say (x=7 or x=7*y) Here y can be any positive integer (y=1 or greater)

*Now from Point 3 we know that its a square. that means it has some form of (x= 7^2) or (x=7^(2) .y) or (x=7^2 . z^2). where y = z^2. 



What I am trying to say here is 


*

*7 is a prime number. 

*x is a square and divisible by 7.


This forces the number x to be a product of 7^2. Which makes it divisible by 49. because 49 is 7^2.  
