Suppose we have two co prime numbers a and b. Then it is always possible to form any number greater than or equal to a*b - a - b +1 by using the given co primes only that is  ax + by where x and y will be non negative integers. I can't seem to find the logic behind it? How it is happening and why?
122619
e.g. given two numbers $5$ and $6$ the maximum number is $19$, as after $19$ each number can be formed using equation $5n+6m$ by putting different (non-negative) values for $n$ and $m$.     Such number is not possible for $4$ &amp; $6$.     I am not sure how to go-forward. The only thing I have till now is that the number is possible only if the given numbers are pseudo-prime. Any help is appreciated.
52713
The problem statement is: for $a,b$ coprime, prove that there exists positive integers $x,y$ such that $ax-by=1$.  The question is form Arthur Egnels' problem solving text, and from the chapter on the pigeonhole principle.    My question: His proof begins by generating a list $a,\ldots,a(b-1)$, where he then points out that each element in this list(which proceeds sequentially) doesn't have remainder $0\pmod{b}$.  He then shows that we arrive at a contradiction if we assume that we also don't get remainder $1$ in this list $\bmod b$.  To show the contradiction in this last statement, he first states that we would have positive integers $p,q $ where $ 0&lt;p&lt;q&lt;b$ so that $pa\equiv qa \pmod{b}$.  He then goes on to point out that since we $a$ and $b$ are coprime, we then have that $b| q-p$.  I understand the last part here, but don't see why we get $pa\equiv qa \pmod{b}$.  Thanks in advance
29280
I want to find integers $z=a^2+b^2$ with $\gcd(a,b)=1$.  Clearly all prime numbers of the form $4k+1$ are such numbers, but what other composite numbers also enjoy this property?  e.g. $74=5^2+7^2$, $625=7^2+24^2$.  Please answer.
122936
I know I have to prove both sides here so: $\implies$: Suppose $|a| &lt; b$. Since $|a|$ is a positive number, we know $b$ is a positive number greater than $|a|$:  If $a$ is positive, $a &lt; b$ If $a$ is negative, $a &lt; b$  But for $|a| &lt; b$, it has to be that $a &gt; -b$. $a &lt; b$ and $a &gt; -b$, therefore $–b &lt; a &lt; b$ is true. $\impliedby$: Suppose $-b &lt; a &lt; b$.  If $a$ is positive, then $b$ is a greater positive number, and $-b$ would be less than $a$ If $a$ is negative, then $b$ can only be a greater positive number for $-b &lt; a$ to hold  So, $b$ is a greater positive integer than $a$. Therefore, it must be that $|a| &lt; b$. This was what I came up with but it seems convoluted and kind of forced with the cases. Any suggestions?
26965
Prove or disprove: For a prime number p, there is a positive integer n such that $p+1 $= $n^2$ if and only if $ p = 3.$  first im trying to proof if p is prime number there is positive integer n such that $p+1=n^2$ implies $p=3$  suppose there is positive n integer such that $p+1=n^2$ and prove that $p=3$. $p=n^2-1=(n+1)(n-1)$ since $p$ is prime $n-1=1$,$ n = 2 $ $p+1=4$ $p=3$ hence it is proved that $p=3$  now if $p=3$ implies there is positive integer n such that $p+1=n^2 $  $p=3$ and p is prime  $p+1=n^2$, $4=n^2$, $n2$ or $n=-2$ hence it is proved that there is positive integer $n$ such that $p+1=n^2 $  is my proof correct? and also for prime number can i write $n+1=1$ instead $n-1=1$?
68102
I have thought a lot but am failing to arrive at anything encouraging.   First try: If this is to be proved by contradiction, then I start with the assumption that let $n$ be a number which is a sum of two numbers, of which at least one is prime. This gives $n = p + c$, where $p$ is the prime number and $c$ is the composite number. Also, any composite number can be written as a product of primes. So I can say, $n = p + p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$. From this, I get $n - p = p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$, but I have no clue what to do next.  Second try: For an instant let me forget about contradiction. Since $n &gt; 11$, I can say that $n \geq 12$. This means that either $p \geq 6$ or $c \geq 6$. Again I'm not sure what to do next.  Finally, consider that the number 20 can be expressed in three different ways: $17+3$ (both prime), $16+4$ (both composite), and $18+2$ (one prime and one composite). This makes me wonder what we are trying to prove.  The textbook contains a hint, "Can all three of $n-4$, $n-6$, $n-8$ be prime?", but I'm sure what's so special about $4, 6, 8$ here.
89037
Let $n$ be an integer such that for any integer $d$, if $d | n$ then $d + 1 | n + 1$.   Show that $n$ is a prime number or equal to $1$  My workout...  Suppose $n = xy$ for some postive integers $x$ and $y. x|n$, and hence we assume without loss of generality $x + 1|n + 1.$   Then $x|xy + 1$. However, $xy + x$ is divisible by $x$.  This implies $x − 1$ is divisible by $x$, which is an impossibility if $x$ is not equal to $1$.  Therefore whenever $n$ is decomposed into two factors, one must turn out to be $1$. Hence $n$ must be a prime.  I am looking for any other method which solves this quickly and more quickly...
38510
Prove: If a prime number $p\in \mathbb N$ is from the form $p=4k+3,k\in \mathbb N$, then its also a prime number in $\mathbb Z[i]$,i.e. if $p|(z_1\cdot z_2)$ then $p|z_1$ or $p|z_2$.  I dont have any idea how to solve it, so I am looking for some hints.  Thanks in advance :)
49434
If $\gcd(a,b)=1$, $\gcd(a,y)=1$ and $\gcd(b,x)=1$ then prove that $ax+by$ is prime to $ab$.  I tried assuming Diophantine Equations for all the relations and representing everything in terms of $a$, but that didn't lead me anywhere. There must be an intuitive and elegant approach right?
81964
Why is $1$ not considered a prime number?  Or, why is the definition of prime numbers given for integers greater than $1$?
121077
For some numbers $n!+1$ is prime, but all such numbers are not prime. For example, $5!+1 = 11\times 11$. The question is this:  Are there infinitely many primes of the form $n!+1$?
85848
Prime numbers are numbers with no factors other than one and itself.  Factors of a number are always lower or equal to than a given number; so, the larger the number is, the larger the pool of "possible factors" that number might have.  So the larger the number, it seems like the less likely the number is to be a prime.  Surely there must be a number where, simply, every number above it has some other factors.  A "critical point" where every number larger than it simply will always have some factors other than one and itself.  Has there been any research as to finding this critical point, or has it been proven not to exist?  That for any $n$ there is always guaranteed to be a number higher than $n$ that has no factors other than one and itself?
112393
Prove that if $p$ is a prime of the form $3n+1$, $p$ is also of the form $6m+1$.  I guess one could use the division algorithm for this; but, what's the best method?
77147
The proofs given on other sites weren't that clear and used different methods that I have yet to learn.    Prove that there are an infinite number of primes of the form 6n+1.  The hint that was given was: Let p = p1, p2, ..., pk + 1, where p1 = 2, p2 = 3,...pk are the first k primes.  Show that p is prime.  (p1 means p sub 1, p2 means p sub 2, and pk is p sub k.  Wasn't sure how to write it on this.)    Can someone explain this hint on how they came about of p1 = 2, p2=3, etc, and prove this please?    Also, how would the proof change if the form changed?  ("Prove that there are infinite number of primes of the form....")
34522
What are techniques to factor numbers that are the product of two prime numbers? For example, how would we factor $262417$ to get $397\cdot 661$? Would we have to guess that factorization or is there an easier way?
92419
Someone claimed that a number, multiplied by the number after it plus 17 is always prime, and showed several cases. I'm not a complete amateur in Number Theory, and I know that $17*18+17=17*19$, so it does not work for $n\equiv0(\mod17)$ but does it always work for other $n$ values? If not, can someone give me a counter example that I can show that person so they can correct their statement?
9407
If $a_1,a_2,\ldots, a_n$ are distinct primes, and $a_1=2$, and $n&gt;1$, then $a_1a_2\cdots a_n+1$ is of the form $4k+3$.  I have no ideas how to approach this question. Any hints?