# Tagged Questions

Questions on congruences, linear diophantine equations, greatest common divisor, divisibility, etc.

3answers
43 views

2answers
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1answer
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### Probability of choosing a number from the set $\{1,2,\ldots,99\}$ that divided by $5$ has the remainder $2$ and is a multiple of $3$

Good evening to everyone. I have to find the probability of choosing a number from the set {1,2...99} that divided by 5 has the remainder 2 and at the same time it's multiple of 3. I know that the ...
0answers
50 views

### Right angled triangle and Pythagorean triplet

Show that there exists a right angled triangle with rational sides and area $d$ if and only if $x^2,y^2$ and $z^2$ are squares of rational numbers and are in arithmetic progression with common ...
7answers
122 views

### Find all positive integer roots of : $5xy=19x+96y$

Find all positive integer roots of : $5xy=19x+96y$ I tried using decomposition technique but no success...,it seems suitable factorization of this equation is IMPOSSIBLE!! Handy calculations show ...
1answer
45 views

### $n$ divides $2^n-1$ $\implies n=1$

If $n\mid (2^n-1)$, then $n=1$. Somehow I am unsure if I got this right, my 'proof' seems to 'easy'. Can you please give me feedback? So I take a prime divisor $p\mid n$. Then $p\mid (2^n-1)$, ...
3answers
156 views

### Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$

Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$ Obviously $(2,1)$ and $(1,2)$ are two answers. But I was unable to manipulate the equation algebraically giving a useful form for finding all other ...
2answers
52 views

1answer
30 views

### Real polynomials from repunits to repunits ( Putnam 2007 A4) [closed]

Find all polynomials $f$ with real coefficients such that if $n$ is a repunit, then so is $f(n).$ [Note this is a Putnam question, so it is intended to be of easy to middling difficulty as contest ...
2answers
54 views

4answers
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### No solution for the equation $y^{n} = 2x^{n}$ for $n \geq 2$ in positive integers.

Exercise from Nathanson's book. Let $n \geq 2$. Prove that the equation $y^{n}=2x^{n}$ has no solution in positive integers. Attempt: We can write the equation as $y^{n}-x^{n} = x^{n}$. I ...