Let A and B be natural numbers I f there exsits integers M a

Let A and B be natural numbers, I f there exsits integers M and N such that MA+NB = 2, are there necessarily integers X and Y such that XA^2 + YB^2 =2 ? prove your answer

Solution

Let A = 2 and B = 1 (natural numbers)

MA + NB =2

Let M =3 and N = -4

3*2 + (-4)*1 = 6 - 4 => 2

Now

XA^2 + YB^2 =2 ///since we have picked A =2 and B=1

X(2)^2 + Y(1)^2 = 2

4X + Y = 2

now we can pick any values for X and Y which follows this equation

lets pick X =4 then Y =-12

lets pick X=-1 then Y = 6

so there are integers X and Y that follows XA^2 + YB^2 =2