## Tuesday, 1 March 2011

### The Nuclear Age Begins

March 1...    It was on March 1st in 1896 that  Henri Becquerel (that's not him at the top) discovered radioactivity, at least according to Wikipedia.   I imagine if he had looked ahead fifty years he might have kept his discovery a secret.  By a strange twist of fate, it was also on March 1st in 1950 that the US convicted  Klaus Fuchs for spying for the Soviet Union and disclosing top secret information about the atomic bomb.
Then, only four years later, on March 1st of 1954, we detonate the Castle Bravo, a 15-megaton hydrogen bomb on Bikini Atoll in the Pacific Ocean .  The result was the worst radioactive contamination ever caused by the United States.

Okay, not the kind of thing you want to think about?  How about diverting your mind with a quick problem.   Can you solve ...

$x^2 - 2y^2=1 \,$
Just find the first two solutions.  After all, March 1st isn't only about nuclear power; it is also John Pell's birthday.  (Yeah, that's him up top.) He would be 400 today... and the first kid who says, "Almost as old as YOU, Mr. Ballew", will be seriously harmed... don't go there.

Oops, almost forgot. It's World Math Day also...

Anonymous said...

What do you mean "first two solutions"? There is only one solution (an ellipse). If you treat each point on the ellipse as a different solution there is no ordering of them. Or did you mean the first points that occur to you, which are probably (1,0) and (-1,0)?

Anonymous said...

What do you mean "first two solutions"? There is only one solution (an ellipse). If you treat each point on the ellipse as a different solution there is no ordering of them. Or did you mean the first points that occur to you, which are probably (1,0) and (-1,0)?

Pat's Blog said...

Sorry Gas.., I thought it would be clear from Pell's name that I was looking for integer solutions where x and y are both >0... and just as a aside, I think you must have meant a hyperbola rather than ellipse.

For example 3,2 will work and is the smallest pair of non-negative integers (I hope) that work.

I should have been more clear in the question.... mia culpa

Anonymous said...

I checked is 2y^2 + 1 a perfect square for a few numbers before encountering the second solution. Is there a way to directly generate solutions?

Jonathan

Pat's Blog said...

Jonathon,
I'm missing something... if 2y^2 + 1 is a perfect square, then it is a solution...