Problem: Find prime solutions to the equation $p^2+1=q^2+r^2$  I welcome you to post your own solutions as well  I have found a strange solution which I can't understand why it works(or what's the math behind it.) Here it is through examples  Put $r=17$(prime) Now $17^2-1=16\times 18=288=2 \times 144$ (a particular factorization)  $\frac {2+144}{2}=73$  $\frac {144-2}{2}=71$  Solution pair $(p,q,r)=(73,71,17)$  Put $r=23$,  $23^2-1=22\times 24=8\times 66$  $\frac {8+66}{2}=37$  $\frac {66-8}{2}=29$  Solution pair $(37,29,23)$  It works for each prime except for $2,3,5$ Which generate  $(2,2,1),(3,3,1),5,5,1)$  Please explain me how it's working
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Are there integral solutions to the equation $x^2+y^2-z^2=1$?
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Can someone please help us solve this equation. It was on our college entry exam, but no one managed to solve it. The problem is:     Let $p,q,r$ be the roots of $x^3 -x+1=0$.  Then $p^5+q^5+r^5 = ?$   Correct answer was $-5$, but no one managed to solve this problem. We tried different methods, but none came up with a solution. Any help is appreciated.
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We had received some questions on Quadratic equations, But I wasnt able to do one. Here it goes:  Let $a,b$ be natural numbers $a&gt;1$. Also, $p$ is a prime number. If $ax^2+bx+c=p$ for 2 distinct integral values of $x$. Then the number of integral roots of the equation $ax^2+bx+c=2p$ is ? Well I know the answer is $0$, But I am not able to get it properly.
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Alright, I met this example solved in a book, but I want to know how did they come up with the answer:  $$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$  Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.
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For a prime number $p \neq 3$, do we have that$$p = x^2 + xy + y^2$$for some $x$, $y \in \mathbb{Z}$ if and only if$$p \equiv 1 \text{ mod }3?$$I suspect this is true from looking at the example$$7 = 2^2  + 2 \times 1 + 1^2.$$Here is a thought I have so far that might be helpful towards solving this: I know that $$\mathbb{Z}[(1 + \sqrt{-3})/2]$$is a PID. But I am not sure what do from here. Could anybody help?
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Suppose $n = pq$ with $p$ and $q$ distinct odd primes.   Suppose that $\gcd(a,pq)=1$. Prove that if the equation $x^{2} \equiv a\pmod n$ has any solutions, then it has four solutions.   Proof: Suppose $n = pq$ with $p$ and $q$ distinct odd primes and that gcd($a,pq$) = 1. Let us have the equation $x^{2} \equiv a\pmod n$. Then, $x^{2} \equiv a$ (mod $pq$), $p \not = q$. By definition, we can separate the equation into two equations such that $y^{2} \equiv a \equiv b$ (mod $p$) and $z^{2} \equiv a \equiv c$ (mod $q$). Let $g_{p}$ be a primitive root modulo $p$ and $g_{q}$ be a primitive root modulo $q$. Then, $b$ is equal to some power of $g_{p}$ and $c$ is equal to some power of $g_{q}$. With the fact that $b$ has a square root modulo $p$ (i.e. $r^{2} \equiv b$(mod $\ p)$) and $c$ has a square root modulo $q$ (i.e. $t^{2} \equiv c$(mod $\ q)$), there is an even power of $g_{p}$ and of $g_{q}$ such that $b = g_{p}^{2k_{1}}$(mod $\ p)$ and $c = g_{q}^{2k_{2}}$(mod $\ q)$ for some $k_{1}, k_{2} \in Z$. By computing, we have the following:  $r^{2} \equiv b$(mod $\ p)$       $\equiv b^{(p + 1) / 2}($mod $\ p)$       $\equiv (g_{p}^{2k_{1}})^{(p + 1) / 2}($mod $\ p)$       $\equiv (g_{p}^{p + 1})^{k_{1}}($mod $\ p)$       $\equiv g_{p}^{2k_{1} + (p - 1)k_{1}}($mod $\ p)$       $\equiv b \cdot g_{p}^{(p - 1)k_{1}}($mod $\ p)$       $\equiv b$ and $t^{2} \equiv c$(mod $\ q)$        $\equiv c^{(q + 1) / 2}($mod $\ q)$        $\equiv (g_{q}^{2k_{2}})^{(q + 1) / 2}($mod $\ q)$        $\equiv (g_{q}^{q + 1})^{k_{2}}($mod $\ q)$        $\equiv g_{q}^{2k_{2} + (q - 1)k_{2}}($mod $\ q)$        $\equiv c \cdot g_{q}^{(q - 1)k_{2}}($mod $\ q)$        $\equiv c$  Hence, $r$ is a square root of a modulo $p$ and $t$ is a square root of a modulo $q$, which means there are two solutions for each $p$ and $q$. Since $p \not = q \in \mathbb{Z}_{n}$, we have isomorphism such that $\mathbb{Z}_{n} \simeq \mathbb{Z}_{p} \cdot \mathbb{Z}_{q}$. Therefore, if the equation $x^{2} \equiv a$ (mod $n$) has any solutions, it must have four solutions. $\blacklozenge$  What do you think about the proof I wrote?
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Problem: If $p\geq 5$ is a prime number, show that $p^2+2$ is composite.  Remarks: Now if one observes that $p$ takes the forms $6k+1$ and $6k+5$, the problem is resolved quite easily. However, if one were to choose other forms say $4k+1$ and $4k+3$ then $p^2$ would be of the form $4p+1$ which would yeild the general form $4p+3$ on addition with $2$. Obviously this does not lead to any concrete conclusion. I was wondering whether there is any specific procedure involved in finding the right quotients (For eg. $6$ and $4$) or is this just a random problem solving trick. Moreover, if anyone could provide some intuition as to why $6$ works would be much appreciated.
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Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$.  This is one of the questions we presented in one session to contest preparation PUTNAM. It turns out that I can't get from the problem. Could someone just give me a hint? (Please, don't give me the answer. Simply, an argument that can help me advance in the problem or theorem might suffice.)
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In the book, "Elementary Number Theory - 6th Edition" written by David M. Burton, I have a question. The problem is  If $p$ and $p^2+8$ are both prime numbers, prove that $p^3+4$ is also prime. (p. 58 problem 3.3 #20)  I don't know how to prove this. Thank you very much if you solve this question.
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Suppose that $5\leq q\leq p$ are both prime. Prove that $24|(p^2-q^2)$.   This is what I got so far. I figured that since $p,q$ are bigger than $5$, there are only odd primes for this conjecture. Then I can rename $p=2m+1,q=2n+1,m,n∈Z$.  We want to show $(p^2-q^2 )=24k,k∈Z$. By substituting $2m+1$ for $p$ and $2n+1$ for $q$, we get  $((2m+1)^2-(2n+1)^2)$. If we expand it, we get $4(m^2+n^2+m-n)$ Q. I think I can make $4(m^2+n^2+m-n)$ look like $24k$. How should I approach ?  And my prof. commented that "I would take this proof in small steps.  Rather than trying initially to prove that $24 | p^2-q^2$, look at the progress you already have.  You proved $4 | p^2-q^2$.  Can you prove 8 is a factor?  What about 3?  Then see if you can put together what you have to get 24 as a factor."  I am stuck here. Any helps?
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Find p is the prime number which $\dfrac{p+1}{2}$ and $\dfrac{p^2+1}{2}$ both are square number.  I do not know how to use "p is prime" assumption given. I just know  $p=7$ is satisfied.  If $\dfrac{p+1}{2}=X^2$ and $\dfrac{p^2+1}{2}=Y^2$ then $(X^2;X^2-1;Y)$ is Pythagorean triple
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Let $p$ be a prime number. Show that the congruence $x^{p-1} \equiv 1 \pmod {p^2}$ has precisely $p-1$ solutions modulo $p^2$  I know the statment above is true for modulo $p$, but how is this imply to modulo $p^2$
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I'm trying to figure out how to prove that if $p$ and $p^2+2$ are prime numbers then $p^3+2$ is a prime number too. Can someone help me please?
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For which primes $p$ is $2^p+1$ divisible by $p$?  What I have been doing is:  $2^p+1\equiv 0\pmod p$  $2^p\equiv -1\pmod p$  Then by Fermat's Theorem, we get  $2^p\equiv 2\pmod p$  This shows $p=1,3$.  I feel like this is not right though...
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Inspired by        Can we solve the equation $$1=\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\cdots +\frac{1}{a_n}$$ where $a_1,a_2,a_3,\cdots ,a_n$ are distinct prime powers (primes are allowed) ?   I tried a modified greedy-algorithm with different starting vectors and brute force with $6$ entris with limit $200$, but did not find a representation.
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Use the modulo version of the quadratic formula and Euler's criterion to decide if the following has a solution or not.  $2x^2+5x+8 \equiv 0\pmod{37}$  I'm not sure how I would use what was being asked of me to decide if this has a solution or not, but I have came to the conclusion that it does not, because I plugged in every possible answer $0-36$, and none of them were congruent to zero.  The way people have answered the question, isn't exactly the way I need.  I have made some progress, but I'm not exactly sure how to finish:  $2x^2+5x+8 \equiv 0\pmod{37}$  $x^2+21x+4\equiv0\pmod{37}$  $x^2+58x+841\equiv837\pmod{37}$  $(x+29)^2\equiv23\pmod{37}$  I know that if $p$ is an odd prime and $p$ doesn't divide $a$, then $x^2\equiv a\pmod{p}$ has a solution or no solution depending on whether $a^\frac{p-1}{2}\equiv 1  \space\space \text{or}\space -1\pmod{p}$.  This is where I get stuck.  Can someone help me from here?  I have a feeling this problem has no solutions, so could someone also explain to me what to do if a problem like this did have solutions.
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If $p$ and $q$ are distinct primes , prove that $$p^{q-1}+q^{p-1} \equiv 1 \pmod {pq}$$     Using Fermat's Theorem we can get $$p^{q-1} \equiv 1 \pmod q, \qquad q^{p-1} \equiv 1 \pmod p.$$  After that I have no idea at all. Can anyone help me please