Transcript Euler’s Solution for Exponent Four

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Leonard Euler
(1707-1783)
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1723 Euler obtains Master's degree in philosophy for the University of Basel
having compared and contrasted the philosophical ideas of Descartes and
Newton. Begins his study of theology.
1726 completes his studies at the University.
1726 First paper in print.
1727 submitted an entry for the 1727 Grand Prize of the Paris Academy on
the best arrangement of masts on a ship awarded 2nd place.
1726 awarded post in St. Petersburg also serving as a medical lieutenant in
the Russian navy.
1730 became professor of physics at the St. Petersburg Academy of
Science.
1733 promoted to senior chair of mathematics, when Bernoulli left.
1738 and 1740 won the Grand Prize of the Paris Academy.
1741 moved to the Berlin Academy of sciences on the invitation of Frederic
the Great. During the twenty-five years spent in Berlin, Euler wrote around
380 articles
1766 Euler returned to St Petersburg after disputes with Frederic the Great.
1771 Became totally blind. He produced almost half his work totally blind.
1783 Died from a brain hemorrhage.
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Leonard Euler
(1707-1783)
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Studied number theory stimulated by Goldbach and the Bernoullis had in
that topic.
1729, Goldbach asked Euler about Fermat's conjecture that the numbers 2n
+ 1 were always prime. if n is a power of 2. Euler verified this for n = 1, 2, 4,
8 and 16 and, by 1732 that the next case 232 + 1 = 4294967297 is divisible
by 641 and so is not prime.
We owe to Euler the notation f(x) for a function (1734), e for the base of
natural logs (1727), i for the square root of -1 (1777), p for pi, S for
summation (1755), the notation for finite differences Dy and D2y and many
others.
Found a closed form for the sum of the infinite series
z(2) = S (1/n2). Moreover showed that z(4) = p4/90, z(6) = p6/945, z(8) =
p8/9450, z(10) = p10/93555 and z(12) = 691p12/638512875. And z (s) = S
(1/ns) = P (1 - p-s)-1.
Euler also gave the formula eix= cos x + i sin x and ln(-1) = pi
In 1736 Euler published Mechanica which provided a major advance in
mechanics
Also gave cases of n=3 (and n=4) for Fermat (with little mistakes), the Euler
equation for of an inviscid incompressible fluid, investigating the theory of
surfaces and curvature of surfaces, and much, much more …
Leonard Euler
(1707-1783)
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Euler’s Solution for Exponent Four
• He actually proves a little more:
a4+b4=c2
has no solutions.
His proof from Elements of Algebra.
202. Thm: There are neither solutions to x4+y4=z2
nor to x4-y4=z2 except if x=0 or y=0.
203. We may assume x and y are relatively prime.
204. Outline of the strategy. Use infinite descent.
I.e. from any solution produce a smaller solution.
But there are no solutions for small numbers.
Leonard Euler’s :Elements of Algebra
205. The case x4+y4=z2 (*)
I. If x,y relatively prime, then (A) either both odd or (B) one is odd and the other is
even.
II. (A) is not possible: An odd square is of the form 4n+1 (If k=2m+1,k2=4(m2+m)+1).
So the sum of two odd squares is of the form 4n+2. This means it is divisible by 2
and not by 4, so it is not a square. So the sum cannot be a square. But 4th powers
are also squares, so the equation cannot hold.
III. (B) If (x,y,z) is a solution (x2)2+(y2)2=z2, then there are p,q such that
x2=p2-q2 and y2=2pq.
IV. Moreover, y is even and x is odd. Then p is odd and q is even. Proof: First since
x2=p2-q2, either p or q has to be odd and the other even. Also p cannot be even,
since then p2-q2 would be of the form 4n-1 or 4n+3 and cannot be a square. (If p=2k
and q2=4m+1, the p2-q2=4(k2-m)-1).
V. Now (x,q,p) is another Pythagorean triple x2=p2-q2, so there are r, s s.t.
p=r2+s2, q=2rs, x=r2-s2.
VI. 2pq=4rs(r2+s2)=y2 so 4rs(r2+s2) must be a square. Also r, s, and r2+s2 have no
common prime factors (why?).
VII. If the statement of VI holds all the factors -r,s, r2+s2- must be squares (why?).
So there are t, u such that r=t2, s = u2. Also r2+s2=t4+u4 is a square and hence
(t,u, (t4+u4)1/2) is a solution of (*).
Since x2=p2-q2=(t4-u4)2, y=t2u2(t4-u4) it follows that x,y>t,u.
So that if (x,y) yields a solution we have another solution (t,u) which is smaller.
VIII. Repeat step VII to obtain smaller and smaller. But there are no small solutions
(except the ones with zeros).
IX. But the smaller solutions from above are also non-zero. This yields a contradiction.
Euler and the case n=3
Consider x3+y3=z3 with relative prime (x,y,z)
1.
Exactly one of the integers is even.
2.
Say z is even and so x,y odd. Then x+y=2p and x-y=2q are both even.
3.
Factorize x3+y3=(x+y)(x2-xy+y2).
4.
Then by inserting x=p+q, y=p-q one finds that p,q have opposite parity
are relatively prime and 2p(p2+3q2) has to be a cube.
5.
The same conclusion is also true if z is odd.
6.
(A) 2p and (p2+3q2) are relatively prime then they have to be cubes and
Euler argues that there exist (a,b) such that p=a3-9ab2, q=3a2b-3b3.
7.
Factor to obtain 2p=2a(a-3b)(a+3b) is a cube and show that the factors
are relatively prime making them all cubes. Say 2a=a3,a-3b=b3,a+3b=g3.
Then (b,g,a) is a new smaller solution.
8.
This also works in a variation if 2p and (p2+3q2) are not relatively prime.
The problem is in step 6. He does not show that this is the only way for
(p2+3q2) to be a cube.