Will assuming the existence of a solution ever lead to a contradiction? I'm reading Manfredo Do Carmo's differential geometry book. In section 1-7, he discusses the "Isoperimetric Inequality" which is related to the question of what 2-dimensional shape maximizes the enclosed area for a closed curve of constant length. He mentions that 

A satisfactory proof of the fact that the circle is a solution to the isoperimetric problem took, however, a long time to appear. The main reason seems to be that the earliest proofs assumed that a solution should exist. It was only in 1870 that K. Weierstrass pointed out that many similar questions did not have solutions.

This line of reasoning would suggest that assuming the existence of a solution might lead to a contradiction (such as an apparent solution that is not in fact valid). Is this actually a problem?
Are there any problems that produce invalid solutions under the (flawed) assumption that a solution exists at all? If so, what is an example and how does it differ from the statement of the isoperimetric problem?
 A: Just the first thing that came to my mind... assume $A=\sum_{n=0}^{\infty}2^n $ exists, it is very easy to find $A $: note $A=1+2\sum_{n=0}^{\infty}2^n =1+2A $, so $A=-1$.
Of course, this is all wrong precisely because $A $ does not exist.
A: Here is a "joke" due to Perron showing that assuming the existence of a solution is not always a very good idea: 
Theorem. $1$ is the largest positive integer.
Proof. For any integer that is not $1$, there is a method to obtain a larger number (namely, taking the square). Therefore $1$ is the largest integer. $\square$
A good source is V. Blåsjö, The isoperimetric problem, Amer. Math. Monthly 112 (2005), 526-566.
A: Let the natural number $x$ be the solution to $x=x+1$.  Then (picks random contradiction out of a hat, as allowed by the principle of explosion (obligatory xkcd)), $4$ is prime.  Proof:  $3$ is known prime, since all smaller positive integers are either units or primes not dividing $3$.  Also, since $3-x = 4-(x+1) = 4-(x)$, adding $x$ to both sides gives $3=4$.  Since primeness is preserved by equality, $4$ is prime.
A: The danger Weierstrass points out is similar to the issue that comes up in the following problem:

What is the minimum value of $x^3-3x$ on $\mathbb R$?

You can easily show with calculus that the only local minimum of this function is $x=1$. Therefore, if the function has a minimum, it must be at $x=1$. However, of course, this function has no minimum, so this reasoning has failed.
In the context of the isoperimetric inequality, the fear would be that there could be shapes with the same perimeter as a circle, but greater area - but perhaps as we add more area, the shapes get increasingly weird and we can't somehow take a limit to get a shape of maximum area.
Really, this should be thought of more as a continuity and compactness issue than an existence issue - we are looking for some way to control the behavior of a function (the area) on a set (the shapes of a fixed perimeter) and know that the circle is the only candidate for a minimum. We would like to say that this implies that every other such shape has less area than the circle, but this requires that we know something about the function and its domain to rule out possibilities like the $x^3-3x$ example.
A: In the general case, does not any proof that uses Reductio ad absurdum meet the criteria requested by the OP?
In particular, the case that I remember best is from high school, that being the proof that $\sqrt 2$ is irrational.  This starts by assuming there exist some $a$ and $b$ that are both integers, have no common divisor, and that the following equation is true: $a / b = \sqrt 2$.
If I remember correctly (it's been over 40 years) the proof proceeds to show that if the above equation is true, $a$ and $b$ do have a common divisor, thus leading to a contradiction.
Hence in this case, assuming the existence of a solution to the problem of finding integers $a$ and $b$ that meet all the criteria of the second paragraph does indeed lead to a contradiction.
A: Let's find the maximum of the the function $f(x) = x^2$.
Assume it does have a maximum. Since $f$ is differentiable, the maximum must happen at some $x$ where $f'$ vanishes.
$x=0$ is the only solution to the equation $f'(x) = 0$.
Therefore, $f(0) = 0$ is the maximum value of $f$.
A: Literally every problem that has no solution is an example.  Indeed, consider any problem which has no solution, and assume $x$ is a solution to the problem.  Then the problem has no solution, but it also has a solution (namely, $x$).  This is a contradiction, and anything follows from a contradiction.  In particular, for instance, it follows that $x=1$, or $x=2$, or any other conclusion you would like to reach.
Of course, this is somewhat artificial, and there are more natural examples.  But as far as the raw logic is concerned, this is just as valid as a more natural example, and illustrates why assuming something which turns out to be false is always a problem.
(In fact, this argument is essentially the same as Perron's "joke" in John B's answer.  You can think of that example as assuming a solution $x$ exists, and then assuming for a contradiction that $x\neq 1$.  Since no solution exists but $x$ is a solution, we have a contradiction, and therefore $x=1$.)
A: Here is one that actually shook the foundations of mathematics. Assume that there is a set of all sets that are not members of themselves. Symbolically, we define the set $R = \{ x : Set(x) \land x \notin x \}$. Then $R \in R \Leftrightarrow R \notin R$, so either classical logic fails for set membership or we have a contradiction!
A: Suppose $x=\log(x)$ has some solution $x=a$.
$0<a$ is obviously true. Then $0<\log(a)$ as well (as $a= \log (a)$ was assumed), so $1<a$. Then $1<\log(a)$, so $10<a$, so ...
Numerous contradictions unfold from this point onward.
A: Courtesy of NumberPhile:
ASTOUNDING: $1+2+3+4+5+\dots=-1/12$
Sum of Natural Numbers (second proof and extra footage)
One minus one plus one minus one
(These videos are all misleading, and they assume these series can even exist, then proceeding to treat them under algebraic manipulations and rearrangements)
A: A famous example is the existence of a solution for $$p^2 = 2q^2, p,q \in \mathbb Z$$
and $p,q$ share no common prime factors.
It is a rephrasing of a classical proof that $\sqrt{2}$ is irrational by assuming it can be written $\sqrt{2} = p/q$ for two such integers.
One comes to the conclusion that a square integer must have just 1 ( an odd number ) of the prime 2 in it's prime expansion, which is of course impossible, since a square must have each prime occurring an even number of times in it's prime factorization.
A: Assume the existence of 
$$S=\sum_{n=1}^{\infty}(-1)^nn  = -1 +\sum_{n=2}^{\infty}(-1)^nn=-1 +\sum_{n=1}^{\infty}(-1)^{n+1}(n+1) \\=-1 -\sum_{n=1}^{\infty}\left[(-1)^{n}n+(-1)^{n}\right]
= -1 -S -\sum_{n=1}^{\infty}(-1)^n$$
this leads to 
$$S = -\frac{1}{2} - \frac{1}{2}\sum_{n=1}^{\infty}(-1)^n =\color{red}{ -\frac{1}{2} - \frac{1}{2}K}$$
Hence the existence of $S$ entails the existence of $$ K=\sum_{n=1}^{\infty}(-1)^n = -1 +\sum_{n=2}^{\infty}(-1)^n = -1+\sum_{n=1}^{\infty}(-1)^{n+1} = -1-K$$
Then we have $K=-\frac{1}{2}$ and then $S= -\frac{1}{2} - \frac{1}{2}K = -\frac{1}{4} $

Finally we get $$ \color{blue}{ \sum_{n=1}^{\infty}(-1)^nn  =-\frac{1}{4}~~~and ~~~\sum_{n=1}^{\infty}(-1)^n =-\frac{1}{2}}$$
  But blatantly none of the above integrals sum converges in the usual sense.

