Home > Maths, Starters > A puzzle based on 71

A puzzle based on 71

Today I was reading an old carnival-of-mathematics. Carnival 71 to be precise. In the intro, the host Robin Whitty (@theoremoftheday) stated that “71 is the smallest prime that can be expressed as x^2 + xy + 2y^2 where x and y are nonnegative integers. (a mini-puzzle: find x and y giving 71)”.

I find it hard not to attempt puzzles, so I had a little think about it. The puzzle itself would be extremely easy to solve using Trial and Improvement methods, especially with a bit of help from a computer, but that’s no fun so I thought about it logically and decided to attempt it mentally.

I didn’t want to start just plugging numbers in, that’s dull and shows no mathematical thinking, so I considered the problem, it boils down to:

Solve x^2 + xy + 2y^2 = 71

The first thing I noticed was that only one of x and y could be odd and only one even. We were dealing with the sum of 3 terms. The third has a factor of two, so is definitely even. This means that one of x^2 or xy must be even, and the other odd, as if they were both odd or even then the sum would be even, yet 71 is odd. If x and y are both odd then xy and x^2 are both odd, and likewise for even numbers. Going a step further, if x is even, then x^2 and xy are both even, so x must be odd and y must be even.

The next thing I noticed was that x has to be less than 8. As 8^2 is 64, and if you add another 8 you are too high. Likewise, y has to be less than 6, as 2(6^2) is 72.

This left me with x is an odd number less than 8 (so 1,3,5 or 7), and y is an even number less than 6 (2 or 4). Thinking logically, we can rule out x = 5, as the only odd number x^2 + xy that that can end in is 5, and for that to work for the equation 2(y^2) would need to end in six, which would need y^2 to be odd, which it isn’t.

Likewise, we can rule out 1 and 3 as if we put them into our equations we would be way too low, as the biggest 2(y^2) can be is 32. This means x has to be 7. Therefore x^2 is 49. Subbing this in

49 + 7y + 2y^2 = 71

2y^2 + 7y – 22 = 0

(2y + 11)(y – 2) = 0

So y = 2

A nice little mental workout, which I rather enjoyed!

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