# Pierre de Fermat's 410th Birthday Celebrated With Google Math Logo

Aug 17, 2011 • 9:19 am | (37) by | Filed Under Logos

Today is the 410th birthday of Pierre de Fermat, born on August 17, 1601 in Beaumont-de-Lomagne, France. He was a lawyer but was best known for his work in mathematics, as illustrated by today's Google Logo.

In the math field he is known for Number theory, Analytic geometry, Fermat's principle, Probability and Fermat's Last Theorem. He helped form what differential calculus is today.

He died on January 12, 1665 at the age of 63 in Castres, France.

Google being a math company wanted to celebrate his life and contributions with a special Google logo, aka Doodle. Google's alternative text behind the logo reads, "I have discovered a truly marvelous proof of this theorem, which this doodle is too small to contain."

I am not mathematician, but one Google user is asking if the equation written in the logo is correct or actually wrong. A Google Web Search Help thread wrote:

I saw today's image connecting pythagoras theorem in a new way.. You wrote the eqn as X^n+Y^n =\=Z^n [n>2]--------(1) I think it is wrong.....Eqn -1 is correct only if n=/=2 (n not equal to 2) I am i right? because theorem is true only for n=2....

Forum discussion at Google Blogoscoped Forums.

##### Boogiejw1

08/17/2011 03:35 pm

N is equal to 2

##### Charles Yu

08/17/2011 03:35 pm

Of course it can be true for n=1...you just said so yourself. There exists a solution. That's all we need here, since the theorem is only looking for the existence of a solution (x + y = z and n=1). For n > 2, no INTEGER solution exists for x, y, and z. That's the point. You can find solutions in the set of real numbers, but no integer solutions exist.

##### Charles Yu

08/17/2011 03:36 pm

...sorry...should have been more specific...no POSITIVE INTEGER solutions exist. Of course, the trivial case of x = y = z = 0 works for all n other than n=0.

##### Dick

08/17/2011 03:36 pm

you used wikipedia to prove your point.. umm ya wikipedia is notorious for being wrong lol

##### Vanity

08/17/2011 03:36 pm

there should be a condition like x>0, y>0 and z>0

##### Kirkld2011

08/17/2011 03:37 pm

410 " th " birthday?

##### TimmyJimmy71

08/17/2011 03:37 pm

I think you flipped the equation....it's stating that n>2 does not equal.  Your equation above with 1 does equal which is the opposite of the equation.  A question:  any idea what the golden spiral and winged circle inside the G mean?

##### Waltonlakreeshia

08/17/2011 03:38 pm

how far are u in your feild of math

##### Sam

08/17/2011 03:39 pm

Of course it's true for x=1, y=0, and z=1, for any integer n. But that's what us mathematicians call a trivial solution. The equality doesn't hold for any nontrivial x,y,z and any 2

##### that irish guy 727

08/17/2011 03:39 pm

I'm a 5'11, 240 lbs. ex-u.s army geek.  Attempting to solve this equation left me confused at the point [n=\=2]. Does N equal, or not equal, 2? This is some geek sh*t.  And I love it.

##### Sam

08/17/2011 03:40 pm

that last bit should be any integer n greater than 2. :)

##### Jatounit

08/17/2011 03:40 pm

n is an integer

##### Jmbabich

08/17/2011 03:40 pm

it does work for n = 0 or 1.   Anything raised to the 0th power equals 1....

##### D Florez

08/17/2011 03:40 pm

oh man, no wonder you are the upset mexican... >.< and I complain about my childhood.

##### Vanity

08/17/2011 03:41 pm

##### Vanity

08/17/2011 03:42 pm

please tell me if I am wrong otherwise my teacher will punish me

##### Inyourbass

08/17/2011 03:42 pm

If the theorem time=money is correct you all must be government employees, because your pissing money(time) away!

##### Vijay Khasat

08/17/2011 03:42 pm

Not really, this goes without saying. This is a very famous math puzzle for which no one has been able to find a proof in more than few hundred years. Of couurse, n is a + integer. If there is any one who can find  solution, he would be a math genuis of firts kind. Vijay Khasat, M.S., P.E. PS: If you want to solve more intersting math puzzles, contact me. or send some puzzles my way. Have a good day.

##### Doug

08/17/2011 03:42 pm

I understand the equation is true. but is this true for right triangles?

##### Jerzynski

08/17/2011 03:43 pm

Hi,       Yawn cubed.                      Sincerely,                             Matt Jer67nski

##### Maurizio

08/17/2011 03:43 pm

This was Maurizio's FIRST Theorem! Maurizio

##### Charles Yu

08/17/2011 03:43 pm

I can't provide any insight on that one. Sorry. I'm not a mathematician.

##### Vijay Khasat

08/17/2011 03:43 pm

Never, give me an example.

##### Jmbabich

08/17/2011 03:43 pm

I have a B.S in Applied Mathematics from UC Davis and have rigorously studied algebraic number theory....  I just thought citing wikipedia would be less snobby than saying that outright.  :)

##### Oldbil

08/17/2011 03:45 pm

another teacher (a friend) quit last week because her classes and responsibilities kept getting bigger and her paycheck kept getting smaller.dumbing us down.Whats next ? Grinding us down for food ?

##### Bharris

08/17/2011 03:48 pm

Remember, this is just a doodle.  The full statement would be that the equation x^n + y^n = z^n has no non-trivial integer solutions for n>2.  This theorem, which took hundreds of years to prove, is now regarded as true by modern mathematicians.  Wikipedia, flawed as it is, will still guide you to a discussion of the generally accepted proof by Andrew Wiles. To say that the equation is true for n = 1 or n = 2 is not precise.  For example, 1^2 + 2^2 =/= 37^2.  What you mean to say is that there are non-trivial solutions when n=1 or n=2.

##### Jmbabich

08/17/2011 03:48 pm

your case is not applicable... x, y, and z must ALL be POSITIVE INTEGERS.   0 is not a positive integer

##### IWillProveYouWrong

08/17/2011 03:50 pm

It was proved in 1993.  It was proved by andrew wiles, however before this it was considered one of the greatest math problem in history.

##### Bencvalley

08/17/2011 03:54 pm

I love how people want to find errors in a publicly displayed work. "What about zero!" do some wikipedia research first. Also read Google's disclaimer that it is INCOMPLETE. Duuuuuuh, that's not right. This is why I like math and not people.

##### Saintpatrick

08/17/2011 03:56 pm

No, because x and z have to be different numbers considering they are different variables...don't try to argue what has already been proven, especially since you obviously don't know much about math.

08/17/2011 04:00 pm

Google Doodles are great! I just found this all google doodle website. http://www.goologos.com

##### Vijay Khasat

08/17/2011 04:05 pm

With due respect, it will not work for zero. Show me, how with an example. God cannot help you in this case.

##### Jeremy Denslinger

08/17/2011 04:31 pm

x = 1 y = 2 z = 3 n = 3 (1^3)+(2^3)≠(3^3)     1    +    8    ≠   27 x = 24 y = 25 z = 26 n = 14 (24^14)+(25^14)≠(26^14)(2.10357201 × 10^19) + (3.7252903 × 10^19)  ≠  (6.45099747 × 10^19) x = 81 y = 26 z = 30 n = 2.26 (81^2.26)+(26^2.26)≠(30^2.26) 20567.246 +1 577.02932 ≠ 2 179.18467 These were just simple examples. I'm sure if someone got real tricky, going to depths to find very specific numbers, the equation would break down... but yeah, as it stands, x^n+y^n≠z^n Oh, wait, I just broke the equation x = 1 y = 2 z = 2.080083823 n= 3 (1^3)+(2^3)≠(2.080083823^3)     1    +    8    ≠             9 ^^^^^^^^^^^^^^^^^^^^^^ !!!!!!!!!!!!!!!!!FALSE!!!!!!!!!!!!!!!!

##### Jrsousa2

08/17/2011 09:50 pm

Great inequation. Too bad the proof is not within reach since it's sooo long.

##### Yirmin

08/22/2011 07:23 pm

And a few hundred years ago it was "proved" that the world was flat... because someone said it was. The fact is this equation is only true when it is constrained by several different points... to me it is a trivial equation... not really worth the time it takes to write.

##### VIJAY KHASAT

08/23/2011 01:05 am

With due respect, I ( a dumb engineer) dare say that 0 WILL NOT WORK. If you think it will, please give me an expample.  One (1)  would work. Ex: 1^1 + 2^1 = 3^1. Am I missing some thing? Vijay Khasat

##### VIJAY KHASAT

08/23/2011 01:09 am

Sir or Madam, I think you are missing the point. This is a century old vary famous math problem. Nobody has been able to come up a proof, and nobody has been able to prove otherwise. With due respect, it has nothing to do with the old belief that the earth was flat. Vijay Khasat