PI Day (3-14) 2019 Puzzle

This puzzle showed up in my twitter feed today, so I thought I’d post the solution as it makes use of both algebra and deductive reasoning. I’ll point out that the problem statement as seen in initial screen isn’t complete, and the solution he gives is to a somewhat different problem. What I solved is this:

Given the equation

\sqrt{PIE} = \sqrt{PI} + E

Where P, I, and E, are integers from 0 to 9, and P cannot be 0, find values for each of these that solve this equation.

The first thing I do is get all of the radicals on the same side …

\sqrt{PIE} - \sqrt{PI}  =  E

then square both sides and gather like terms, pulling out the common factor, PI …

PIE - 2PI\sqrt{E} + PI  =  E^2

PI(E - 2\sqrt{E} + 1)  =  E^2

Since we’re only considering integers, and E is the only term under the radical, our choices for it are limited to 0, 1, 4, and 9.

We can eliminate 0 right away. The only way the equation makes with E=0 is if P or I are also zero. P can’t be (by definition) and I can’t be since we’re not allowed to have both E and I set to zero.

Setting E = 1 makes the term in parentheses 0, which results in 0 = 1, which is of course absurd.

With E=9 we end up with 4 in the parenthetical, making the whole equation

4PI = 81

Since an product with an even factor must also be even, this rules out 9.

 

That leaves us with 4.  Plugging that in gives us PI = 16 which is entirely plausible. 

In fact, there are only two numbers left in our range that solve this: 2 and 8.  It doesn’t matter which you assign to P or I.  The math still works out.

Plugging these back in to the original equation, we see that

\sqrt{(2)(8)(4)} = \sqrt{(2)(8)} + 4

\sqrt{64} = \sqrt{16} + 4

8 = 4 + 4

The Original Puzzle

The puzzle I linked to at the top has one additional constraint: PIE is a three digit number, and PI is a two digit number. If you watch the video, the solution is almost pure deductive reasoning with little if any algebra at all. Oh, and the solution itself is completely different. In fact, one of the solution elements that doesn’t work here, does work in his version of the problem.