For the first Math Trick of the Week (MTW), which will be uploaded every 23:30 Wednesday GMT time (as near that time as I can make), is going to be on... SFFT.
WHAT DOES SFFT STAND FOR?
SFFT stands for: Simon's Favorite Factoring Trick, which was popularized by AoPS (Art of Porblem Solving), in one of their videos.
WHAT IS SFFT?
To explain it, let us look at an example: "Solve for all integer pairs (x,y) in this: xy+x+y=0". At first glance, this problem seems hard. Then, we might observe this solution: (0,0). But, is that the only one? If we can find another, how can we prove there are only these pairs?
Method 1 (Works in general, but slow)
xy+y=-x
(x+1)y=-x
y=-x/(x+1)
So, when is y an integer given x is an integer? Well, after a bit of deliberation, it is clear that only x=0 and x=-2 work, so the pairs are (0, 0) and (-2, -2). (Took me 1:34 mins using this method)
Method 2 (Fast, SFFT)
xy+y=-x
(x+1)y=-x
(x+1)y-1=-(x+1)
(x+1)(y+1)=1
So, (x+1, y+1)=(1, 1) or (-1, -1). So, (x,y)=(0, 0) or (-2, -2)
What is SFFT in general?
axy+bx+cy=d
xy+(b/a)x+(c/a)y=d/a
(y+b/a)x+(c/a)y=d/a
The next step, is to make y+b/a.
(y+b/a)x+(c/a)y+bc/(a^2)=d/a+bc/(a^2)
(y+b/a)x+(y+b/a)(c/a)=d/a+bc/(a^2)=(da+bc)/(a^2)
(y+b/a)(x+c/a)=(da+bc)/(a^2)
(ay+b)(ax+c)=da+bc.
You now find factors of the right hand side, and then you can find (x, y).
Conclusion
SFFT is a powerful trick that solves diophantine equations of the form axy+bx+cy=d. I hope you understand this trick now.
For the first Math Trick of the Week (MTW), which will be uploaded every 23:30 Wednesday GMT time (as near that time as I can make), is going to be on... SFFT.
WHAT DOES SFFT STAND FOR?
SFFT stands for: Simon's Favorite Factoring Trick, which was popularized by AoPS (Art of Porblem Solving), in one of their videos.
WHAT IS SFFT?
To explain it, let us look at an example: "Solve for all integer pairs (x,y) in this: xy+x+y=0". At first glance, this problem seems hard. Then, we might observe this solution: (0,0). But, is that the only one? If we can find another, how can we prove there are only these pairs?
Method 1 (Works in general, but slow)
xy+y=-x
(x+1)y=-x
y=-x/(x+1)
So, when is y an integer given x is an integer? Well, after a bit of deliberation, it is clear that only x=0 and x=-2 work, so the pairs are (0, 0) and (-2, -2). (Took me 1:34 mins using this method)
Method 2 (Fast, SFFT)
xy+y=-x
(x+1)y=-x
(x+1)y-1=-(x+1)
(x+1)(y+1)=1
So, (x+1, y+1)=(1, 1) or (-1, -1). So, (x,y)=(0, 0) or (-2, -2)
What is SFFT in general?
axy+bx+cy=d
xy+(b/a)x+(c/a)y=d/a
(y+b/a)x+(c/a)y=d/a
The next step, is to make y+b/a.
(y+b/a)x+(c/a)y+bc/(a^2)=d/a+bc/(a^2)
(y+b/a)x+(y+b/a)(c/a)=d/a+bc/(a^2)=(da+bc)/(a^2)
(y+b/a)(x+c/a)=(da+bc)/(a^2)
(ay+b)(ax+c)=da+bc.
You now find factors of the right hand side, and then you can find (x, y).
Conclusion
SFFT is a powerful trick that solves diophantine equations of the form axy+bx+cy=d. I hope you understand this trick now.
THE END