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Euler creates a structure/identity (which are now generalised and known as Newton's identities) to solve a depressed quartic equation, where the coefficients are essentialy that of a cubic. We know by Newton's identies we can extended this pattern for a depressed quintic in terms of elementary symmetric functions. It should look something like this $$p^5+q^5+u^5+v^5= x^5 -5(pq+pu+pv+qu+qv+uv)x^3.... $$ where $x = p+q+u+v$. The question is: what was eulers attempts/insights in all of this?

Sebastiano
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user781074
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  • Are you saying we can turn a depressed quintic into an equivalent quartic equation? If so please provide more detail. – anon May 31 '20 at 07:26
  • I said almost equivalent. Look at the quartic equation coeffiencints of (x+p)(x+q)(x+u)(x+v) but instead the additive coefficient (p + q + v + u) will be ( p^5 +q^5 + u ^5 + v^5) for the quintic structure because x = p+q+u+v . Hence almost similar to a quartic equation – user781074 May 31 '20 at 07:34
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    And so you end up with another quintic. That's why it leads nowhere in general. – Conifold May 31 '20 at 07:43
  • That's obvious. I'm just curious of eulers attempt . Know any mathematicians that made good attempts of trying to find a solution by radicals? (Even tho we know they'll fall short) – user781074 May 31 '20 at 14:08
  • You may want to look at Lagrange's work, which followed Euler's, see The Theory of Equations in the 18th Century. After Lagrange it became clear that old substitution tricks will not work, soon it was proved (Ruffini, Abel) that nothing will. In the 19th century people focused on looking beyond radicals, and solutions were found using elliptic functions, see Thomae's formula. – Conifold Jun 01 '20 at 07:58
  • If you're looking for the historical perspective here, you might ask at the History of Science and Mathematics StackExchange. – Blue Jun 06 '20 at 14:29

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