It is required to prove that for positive integers $a$,$b$ if $$1+ab|a^2+b^2$$ then the integer  $$\frac{a^2+b^2}{1+ab}$$  is a perfect square. The proof is easily beyond me as I have only recently started studying number theory or maybe I'm not intelligent enough. Help will be highly appreciated.
1879
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 root is determined in two different ways from the expansion $(x-r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1 r_2$.)  Now suppose we have $a,b$ such that $k$ is an integer but not a square, by the investigation about roots we have that the second root is a nonzero integer since $k,b,a$ are integers and $k \not = b^2$, futhermore it is positive which is easily seen from its defining equation.  WLOG assume $a \ge b$ so that the second root is strictly smaller than $a$. This leads to a decent, replacing $a$ with the second root.    Proof 2 (Don Zagier): Apply reduction theory (specifically, Sätze 1 and 2 of Section 13 of my book on quadratic fields) to the quadratic form $x^2 + kxy + y^2$, which is the unique reduced quadratic form in its equivalence class.    Note that Proof 2 is pretty much the same as Proof 1 when written out in explicit detail, but I could not read Zagier's book because I cannot read German.  I would like to know more approaches to this and other alternative proofs of this result if possible! Thanks in advance.  I would also be interested in related problems (especially easier ones of a similar nature) and texts which cover the reduction theory in English.
23022
If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$.   I know how to do the direct proof, but in this case, I want to try proving it by contradiction. I have tried manipulating the inequality $a + b &lt; 2 \sqrt{ab}$ after making the assumption that $a,b &gt;0$ to get a contadiction $a,b \leq 0$.  $\begin{align} a + b &amp;&lt; 2 \sqrt{ab} \\a^2-2ab+b^2 &amp;&lt; 0 \\(a-b)^2 &amp;&lt; 0\end{align}$  How do I show that $a,b \leq 0$?
103821
I saw this problem recently and found an elegant solution to it, and was curious to see if anybody would think of something else. Nice solutions to nice problems are fun to see!  Problem: Prove that, for all non-negative integers $k$, there exist integers $a$, $b$ such that $a^2+b^2=5^k$. Bonus: Prove that there exist such $a$ and $b$ that furthermore satisfy $(a,b)=1$.
48501
Is every sufficiently large positive integer $A$ of the form $ab + ac + bc + 1$   where $a,b,c$ are some positive integers larger than some given positive integer $d$ ?  How large is sufficiently large , in other words how do we know our number $A$ is sufficiently large ( as a function of $d$ ) ?    Update :  I read the following about positive integers that are not of the form $ ab + ac + bc $:    And I quote about that sequence above : " Note that for n in this sequence, n+1 is either a prime, twice a prime, the square of a prime, 8 or 16 "  This is why I considered the weird +1 in $ab + ac + bc + 1$  So composites larger than 16 not a prime , twice a prime or the square of a prime are of the form $ab + ac + bc + 1$.  And I wonder why !  I guess that relates strongly to the Original posted problem. Maybe I should have mentioned that before , sorry.  Will Jagy's answer gave the same OEIS link which I already knew. This also explains some of the comments.  Im no expert in genus theory or Galois theory but I assume this has a simple answer.
95318
I'm trying to show that if $\alpha$ is the square root of a non-perfect square $d$, then for any rational $p/q$ there's a positive real $x$ such that $|\alpha - p/q| &gt; x/q^2$. I'm having trouble starting on the problem, could someone give a hint? Thanks
10320
If $a$ and $b$ are two relatively prime positive integers such that $ab$ is a square, then $a$ and $b$ are squares.   I need to prove this statement, so I would like someone to critique my proof. Thanks  Since $ab$ is a square, the exponent of every prime in the prime factorization of $ab$ must be even.   Since $a$ and $b$ are coprime, they share no prime factors.   Therefore, the exponent of every prime in the factorization of $a$ (and $b$) are even, which means $a$ and $b$ are squares.
112624
Find all primes $p$ and $q$ such that $p^2 + 7pq + q^2$ is a perfect square.  One obvious solution is $p = q$ and under such a situation all primes $p$ and $q$ will satisfy.  Further if $p \neq q$ then we can assume without the loss of generality that $p &gt; q$. Assuming this and that there exists at least one such perfect square I have tried to show some contradiction modulo $4$ as any odd perfect square leaves a remainder of $1$ when divided by $4$, but it is not working. However I firmly believe that $p = q$ is the only solution, but I have failed to prove this.
72873
Find all prime numbers $p$ such that $p!+p$ is a perfect square.  I think the only ones are when $p!+p=p^2$, i.e. $p\in \{2,3\}$.  Any ideas at all?
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!
34795
Does anyone know an "elementary" proof of the following theorem?  Let $k \neq 0$ be a rational integer. Then $k$ admits a square root in $\mathbb{Z}_2$ if $k = 4^a (8b+1)$ for some $a \in \mathbb{N}$, $b \in \mathbb{Z}$.  About $p$-adic numbers I don't know anything more sophisticated than Hensel lemma. Thank you!
49227
Prove that for each $n$, there exist $n$ consecutive integers, each of which is divisible by a perfect square larger than $1$. Can any one help with this? I was stuck because I know how to solve for a particular $n$ value, but I was helpless in the generalized form.
62095
If $x$ and $y$ are positive integers, then when is $2^x+3^y$ a perfect square?   I tried this question a lot but failed. I tried dividing cases into when $x,y$ are even/odd, but still have no idea what to do when they are both odd. I tried looking into when is  $2^x+3^y$ is of the form $6k+1$ and so on .. but couldn't succeed. Can anyone help?
89623
Let $d$ be any positive integer not equal to $2, 5,$ or $ 13$. Then can we always find distinct $a, b \in \{2, 5, 13, d\}$ such that $ab − 1$ is not a perfect square ?
13594
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 $n-1$. What is the answer for a general value of $n$?  (And is it possible, to prove without Bertrand's postulate. Because Bertrands postulate is quite a strong result.)
101195
The famous IMO problem 6 states that if $a,b$ are positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $\frac{ a^2 + b^2}{ab + 1 }$ is a perfect square, namely, $gcd(a,b)^2$.  How about a modification of this problem:    If $a,b$ are (strictly) positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $ab + 1$ cannot be a perfect square.    I am looking for a proof of the claim above, or a counter-example.  One possible approach towards a proof is the following:   Suppose there is such a pair $(a,b)$ as above, then by the famous IMO problem 6 from 1988, $\frac{ a^2 + b^2}{ab + 1 } = g^2$ where $g = gcd(a,b)$. Since $ab + 1$ is a perfect square, $a^2 + b^2 = c^2$ for some integer $c$. So that $(a,b,c)$ is a Pythagorean tripple, therefore there exists positive integers $n,m,l$ with $n$ coprime to $m$, such that $a = l(n^2 - m^2)$ and $b = 2lnm$ - by Euclids formula.  Then by plugging in $a$ and $b$ in terms of $n,m,l$ into the original equation, and solving for $l$, it is possible to obtain the following:  There exists positive coprime integers $n,m$ such that   $\frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)}$ is a perfect square.  If we put the quotient above into a program, as in this python code snipet:  N = 1000  for n in range(1,N):     for m in range(n+1, N):         A = (n*n + m*m + 1)*(n*n + m*m - 1)         B = 2*m*n*(n+m)*(m-n)         if A % B == 0:             print(A/B)   The quotient always is 2, regardless of whether or not $n$ and $m$ are co prime! So if the very strong implication   $2mn(n+m)(n-m) \mid (n^2 + m^2 + 1)(n^2 + m^2 - 1) \implies \frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)} = 2$    holds, then the original problem will be solved.
34817
So, I decided to work with mod $8$ to help develop some intuition on how to generalize the proof. I noticed that taking $a^2$ to clearly be a perfect square, $a^2$ is always congruent to $0,1,-4,4 \mod 8$, at least for a good finite set of cases. Also, I played around with solutions to $3^m + 3^n + 1 = b$ and seen again for a finite set of cases with $m,n$ varying these were always, $b$ is congruent to $-1,5,-5,3$. Now, I see that well only perfect squares produce a set $\{0,1,-4,4\} \mod 8$ and none of the cases I tried to $m,n$ gave me any of the numbers in this set. And now I'm stuck.
99756
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that    $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$     My work: $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}$$ Now, we have $a^2+\dfrac{1}{a^2}\ge 2$ and $b^2+\dfrac{1}{b^2}\ge 2$. Here, I am stuck, I cannot use the information provided, $a+b=1$ to any use. Please help!