I am trying to prove or disprove that if $x$ is even, then $x + 5$ is odd.  This is what I have thus far, but I am stuck:   Assume that the chose variable (x) are in the domain: (x) is an integer Assume the IF part of the statement: (x) is even Prove the THEN part:  Let a be a dummy variable that we assume is an integer. $x=2a+1$ (Definition of an Odd Integer)   Now find $x+5$.  $x+5=2a+1+5$ (Because we added the same value to both sides of the equation)  $x+5=2a+6$   $x+5=2(a+3)$ (Because of the Distributive Property of Multiplication over Addition)  Let $z=a+3$;  $x + 5 = 2z,$ but this is the definition of an even integer; where did I mess up?
121064
Prove that a even number + odd number = odd number  Let $x$ be the even number, let $y$ be the odd number.   From the definition of odd numbers, $y + 1$ is even.   Let:  $$x + y = z$$  Suppose $z$ is even.   $$x + y = z$$  It follows that:  $$x = x + y - y = z - y$$  But I cant think of any other contradiction here. The issue is with the assumption I suppose. Any help is appreciated!
44996
i would like to know how to prove that if $a^2+b^2=c^2$ then   a) $3|a$ or $3|b$  b) $5|a$ or $5|b$ or $5|c$  c) $4|a$ or $4|b$   Also that $a^2+b^2+c^2+1$ can't be divisible by 8  Thank you very much!
64137
A polynomial problem from my old algebra textbook:     $f(x)\in\Bbb Z[x]$ with leading coefficient $1$, $\deg f(x)\ge 1$, and both $f(0)$ and $f(1)$ are odd numbers, prove:   $f(x)$ has no root within $\Bbb Q$   Eisenstein's criterion seems to be of little help here because we know virtually nothing about the coefficients. So I tried some other approach. Let $$f(x)=a_0+a_1x+a_2x^2+\cdots+a_{n-1}x^{n-1}+x^n\in \Bbb{Z}[x]$$ What we have known so far is $$f(0)=a_0\text{ is an odd number}$$ and $$f(1)=a_0+a_1+a_2+\cdots+a_{n-1}+1 \text{ is an odd number}$$ from which we can also conclude that $$f(1)-f(0)=a_1+a_2+\cdots+a_{n-1}+1\text{ is an even number}$$ To be honest I don't know how to proceed now. It isn't until just now that I realized "irreduicible" is not equivalent to "has no roots".. So all my previous attempt is fundamentally wrong.. Can anybody help or drop a hint? Best regards!
89243
I have tried as follows. Please help to double check the proof! Thank you!  Since $x=p/q$ ($p$, $q$ are integers), $(p/q)^2+(p/q)a+b=0$  So, $(p/q)^2=-b-a(p/q)$  then, $p^2=-bq^2-a(p/q)q^2$  and, $p = \dfrac {q(-bq-a(p/q)q)}{p}$  now, it is clear that $q \mid p$, thus $p/q$ must be an integer.
104049
Prove that there does not exist an integer solution for the diophantine equation $x^5 - y^2 = 4$.   It's obvious that $x$ and $y$ are of the same parity. We can also claim that if $x$ is odd, then it is $1 \pmod 4$. Also, if $x$ and $y$ are even, then $y \equiv 2 \pmod 4\text{ since } x^5 \text{ is a multiple of } 32$ and if $y$ were a multiple of $4$ then it would be at a distance of at least $16$ from $x^5$ or be equal to it. These are my observations.  How should I proceed with the proof? Please give hints.
3696
I have a problem in solving my number theory homework. My question is as follows:  Let $p$ be an odd prime. Prove that $a$ is a quadratic residue mod $p$ if and only if the $I_{g}(x)$ (index with respect to any primitive root of $p$) is even.  Please edit my writing. Thanks. Does anyone know where to start? Thank you very much for everything!
90496
Prove by induction that this number is an integer: $$u_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$$ Progress I assumed that it holds for $n$ and I tried to do it for $n+1$ but the algebra gets quite messy and I'm unable to prove that the following term is an integer: $\sqrt{5}((3+\sqrt{5})^n-(3-\sqrt{5})^n)$
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Suppose that $[F(\alpha):F]$ is odd. Show that $F(\alpha)=F(\alpha^{2})$. Generalize this statement if possible.    Clearly, we always have that $F(\alpha^{2}) \subset F(\alpha)$ by definition of the field generated by $F$ and $\alpha$ (assuming that we have a field extension $K/F$ where $\alpha \in K$). However, how can we show the other inclusion? If the degree is odd, then it follows that $\alpha$ will solve a polynomial of odd degree in $K[x]$ (for the coefficients in $F$). What would be a way to proceed from here? (and as well to generalize the statement?)  I was thinking of using the formula: $[F(\alpha):F]=[F(\alpha):F(\alpha^{2})][F(\alpha^{2}):F]$ and maybe proceed from there.   Thanks for the help.
106966
If $c|a$ and $c|b$ then $c|(ax+by)$ where $c,a,b$ are integers  Proof:  Suppose $c|a$ and $c|b$ then we can represent as the following:  $a=cx$ where $x$ is an integer  $b=cy$ where $y$ is an integer  Then adding both:  $a+b=cx+cy=c(x+y)$  therefore $c|a+b$  but how can i show $c|(ax+by)$ ?
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My apologies if this has been asked already, I've searched but couldn't find it...  Let $x$ such that for every $y \in N$, $y^x$ is an integer. Does that necessarily mean that $x$ is an integer?
9603
Prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer.      For clarity: the denominator is the only part being squared.   My thought process:   The numerator is the product of the first $n$ even numbers and the product of the first $n$ odd numbers.   That is, $(2n!) = (2n)(2n-2)(2n-4)\cdots(2n-1)(2n-3)(2n-5).$ In effect, the product of even numbers can be cancelled out with $n!$ resulting in the following quotient:  $$ \frac{(2^n)(2n-1)(2n-3)}{(n!)}\;.$$  To me this looks even thanks to the powers of $2$. But I am not convinced for some reason.   Was this approach to naive?   Sorry for the poor notation, I don't know any coding languages, my apologies.
61123
I need to show the following:  For any integer $x$, $x^2 + 4$ is not divisible by $3$.    I was trying proof by contraposition, but I do not believe that is the most efficient way to go about this.  Can anybody point me in the right direction, please?
30883
If $a$ is an odd integer, prove that $24 \mid a(a^2-1)$.  I proved in my class that $a^2=8k+1$ for some $k$, provided $a$ was odd.  So $$a(a^2-1)=(2k-1)(8k+1-1)=8k(2k-1)$$ but I can't pull a $24$ out of it.
92678
How would I go by proving this statement?  What I did was I tried using Proposition GCD Of One, so that  $(11a+5)x + (2a+1)y = 1$, and $(11x+2y)a + (5x+y) = 1$.  But I have no idea what to do from here. What can I do?
114930
This is a question the prof asked in my university math proofs course. I've tried this for a bit over an hour and I am stuck.   The method I personally chose was proof by cases:  Case 1: n is even:    Thus $n = 2k$ for some k ∈ Z. So $n = 2k = 5a + 2b$  Here I can choose $a=0$, thus $n = 2k = 2b$ Therefore, if n is even there always exists some a,b∈ Z such that $n = 5a + 2b$ and case 1 is true.  Case 2: n is odd:  Thus $n = 2k+1$ for some k ∈ Z. So $n = 2k+1 = 5a + 2b$  This is where I'm stuck, I cant seem to show that the case is always true for when n is odd.  If you could give me a hint in the right direction or complete my method of proof by cases that would be great, or if you had another method of proving it that would be helpful too.  Thanks, John
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What is the logic/thinking process behind deriving an expression for even and odd functions in terms of $f(x)$ and $f(-x)$?  I've been pondering about it for a few hours now, and I'm still not sure how one proceeds from the properties of even and odd functions to derive: $$\begin{align*} E(x) &amp;= \frac{f(x) + f(-x)}{2}\\ O(x) &amp;= \frac{f(x) - f(-x)}{2} \end{align*}$$ What is the logic and thought process from using the respective even and odd properties, $$\begin{align*} f(-x) &amp;= f(x)\\    f(-x) &amp;= -f(x) \end{align*}$$   to derive $E(x)$ and $O(x)$?    The best I get to is: For even: $f(x)-f(-x)=0$ and for odd: $f(x)+f(-x)=0$  Given the definition of $E(x)$ and $O(x)$, it makes a lot of sense (hindsight usually is) but starting from just the properties. Wow, I feel I'm missing something crucial.
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I am trying to prove old test exercises and this one is resisting me.  I know you can believe that I meant $a^ 2 + b ^ 2$ is divisible by 4, but no, in the test it says that 4 is divisible by $a ^ 2+ b^2$.  Suppose that $ a,b \in \mathbb{Z}$  If 4 is divisible by $ a^2 + b^2$, then $a$ and $b$ are not odd.  I started like this:  Suppose that a and b are even, then $a = 2m$ and $b = 2n$ for some $m,n \in \mathbb{Z} $.   Thus  $$ a^2 = (2m)^2 \ \mbox{and}\ \  b^2= (2n)^2$$ $$ a ^2 + b^2= (2m)^2 + (2n)^2$$ $$ a^2 + b^2 = (4 m^2 + 4n^2) = 2(2m^2+2n^2) $$  Now $$2m ^ 2 + 2n^2\in \mathbb{Z} = k$$ Then  $$ a^2+b^2= 2k$$  Is this a correct proof?/ Any better ways to prove this.  Sorry for the lousy English. Thanks.