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Sir Michael Francis Atiyah… is a British-Lebanese mathematician specialising in geometry.

At a hotly-anticipated talk at the Heidelberg Laureate Forum today [Monday 24 September 2018], retired mathematician Michael Atiyah delivered what he claimed was a proof of the Riemann hypothesis, a challenge that has eluded his peers for nearly 160 years.

If *a* = 2, *b* = 1, *c* = 4, then *A* = √(2+1·√ 4) = 2, *B* = √(2−1·√ 4) = 0, *S* = *A* + *B* = 2 ≠ *b*.

bob bundy wrote:

Draw three distinct, parallel lines and another line (a transversal) to cross them all.

But those three lines may not be in one and the same plane.

Let *x* be the number of people enrolled in all three classes. Then:

170−

*x*are enrolled in art and drama but not piano,150−

*x*are enrolled in drama and piano but not art,300−

*x*are enrolled in piano and art but not drama.

Now that you have split the numbers into non-overlapping groups, you can add them up:

Let us apply zetafunc’s method in another way. Dividing by 2 and rearranging gives

Now consider the following table:

Thus we see that outside of {0,±1} the equation has no solution as the LHS and RHS have opposite signs. This proves that there are no solutions other than *x* = 0, ±1.

There are precisely three solutions:

Unfortunately I do not know how to prove it rigorously. Maybe Bob Bundy can come up with more helpful information.

Here’s a YouTube video on tests of divisibility and how to apply them to some math problems.

I found the solution to finding all palindromic four-digit powers particularly interesting.

anonimnystefy wrote:

Can you help me solve this equation?

What do you mean by why is *n* term not 2*n*−1?

Since the thread title is mathematical induction, I take it you want to prove the formula for the sum of the first *n* odd positive integers. To do this, you need to know what the formula is in the first place. If you’re not given the formula, you can make a guess and then proceed to prove it. So, trying the first few sums …

I think you can make a guess as to what the formula is going to be.

I presume *ɑ* and *β* are the roots, which are complex. Since we are going to deal with their powers, best to convert them to cos–sin form:

So the roots are

one of them is *ɑ*, the other is *β*. Since the question doesn’t say which is which, you may as well take your pick. Now use De Moivre’s theorem.

Why can’t they use metric?

Do the long division yourself!

By the way, this thread is better in the Puzzles and Games section.

Zeeshan 01 wrote:

Do that division, given in https://www.mathsisfun.com/long_division.html

Alg Num Theory wrote:

I’m not doing your homework for you. You do it yourself.

In problems like this, the first thing to is to reduce the fraction to its lowest terms:

so you’ll be doing the following long division:

If the denominator ends with one or more 0, you can shift the decimal point in the numerator to get rid of the 0:

giving

I’m not doing your homework for you. You do it yourself.

Go on, then. Use long division.

In that case …

A: Because it has aged with Tyne and is now the worse for Wear!

Bob, I think *r*=0 is also possible for (a).

For (b), I think the answer is all nonnegative integers. Clearly there are infinite geometric sequences with no integers, e.g.

If *r* = 1, add 1 as the first term to the above sequence.

If *r* > 1, then

is an infinite geometric sequence (common ratio 1/*r*) with exactly *r* integer terms (the first *r* of them).