I Solved A Diophantine Equation

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[TedHopp]

After finding the first two solutions, you should assume that y>=4, not 2. That way you don't end up with the silly conclusion that there are no solutions for y>=3 when you already found a solution for y=3. Alternatively, after finding the first solution, just assume y>1 and you can discover the second solution using your same modular arithmetic reasoning.

[angelishify]

For y >= 2 must be m = 1 (x = 2). In this case is y = 3. The other solution is trivial ( x = 1, y = 1).

[williamperez-hernandez]

The explanation at the end is incorrect. The equation for m is m(m+1)=2^(y-2). The only valid power of 2 is exactly 2, so y-2=1 is the only solution for y>=2. Thus y=3

[johns.]

At 4:15, it's clear that either 3^z+1=2 or 3^z -1=2 since no two consecutive even integers are both divisible by 4..

[abdhelal]

But how y = 3 and you arrive to contradiction if y>= 2 ? Sorry I didn’t get it . If you can explain more

[Qermaq]

Here's a circle geometry problem. Inside a circle with radius R are three circles with radii p, q and r. Each circle is tangent to the other three. p+q=R, p-1=q and q-1=r. What is r?

This is a toughie, but spoilery hints below:

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Note that since p+q=R the diameters of p and q are collinear. Let n = p/q. The hard part for me was figuring out r = n^2+n, q = n^2+n+1, and p = n^3+n^2+n. Once you can show that, n and r are simple to find.