Fun with maths

we have  $x^2-xy+y^2 = 13$ we can complete square by addition of some thing from $y^2$ to $x^2-xy$ to get $(x^2-xy + \frac{y^2}{4}) +  \frac...

 We have $ 42 = 2 * 3 * 7$ We need to show that $3^p-2^p=1   \pmod {2}\cdots(1)$ $3^p-2^p=1   \pmod {3}\cdots(2)$ and  $3^p-2^p=1   \pmod {...

 We shall prove if by binomial expansion we have $4^{2n}$ $= (5-1)^{2n}$  $ = \sum_{k=0}^{2n}{2n \choose k}5^{2n-2}(-1)^{k}$ $ = \sum_{k=0}^...

we have $n^2 \equiv 0/1 \pmod 4$ So $m^2 + n^2 \equiv 0/1/2 \mod 4$ So any number of  of the form 4k + 3 cannot be expressed as sum of 2 squ...

Now 7 cannot be multiple of 7 because in that case 7 cannot be factor of either of them Because 7 is prime so using Fermat's little theo...

 We have $2^x = y^2- 1 = (y+1)(y-1)$ As product of y+1 and y-1 is a power of 2 so both are power of 2. y+1 and y-1 one of them is divisible ...

 We have $\frac{1}{a_1} + \frac{1}{a_2}  + \frac{1}{a_3}  + \frac{1}{a_4}  + \frac{1}{a_5}  + \frac{1}{a_6}=$ $\dfrac{a_2a_3a_4a_5a_6+a_1a_3...