Pythagorean Puzzle

[British_Olympics]

A right-angled triangle has sides of length A,B and C where A is the hypotenuse and is an integer. When B and C are multiplied, the result is the square of A divided by one less than the square. When B is divided by C the result is one greater than than A divided by one less than A. What is the value of A ?

Struggling a bit with this one and the second sum seems ambiguous. Does it mean B/C = (A+1)/(A-1) or B/C = 1 + (A/(A-1)) ? I presume the former, but can't see a way to solve the 2 equations which is compatible with (A*A) = (B*B) + (C*C).

[fiction-factory]

It is ambiguous but it reads more like the former.
When I scribbled down a bit of maths using substitution and difference of two squares I got to C= a/(a+1) and for integer values of A then C must always be less than one- yet we know C is an integer. So either I have gone wrong or we are wrong in assuming the former.
Maybe you just need to inspect Pythagorean triples and first find one or more that meets the other conditions

[fiction-factory]

Sorry, I have just read it again and I see that only A is an integer. I will have another think later

[Prudie]

I've jotted this down for a play too but i read the 2nd equation as;

B/C =(A/A-1) + 1

[fiction-factory]

I can see both interpretations. I think the former interpretation looks less likely to lead to a solution so will try your interpretation later, prudie. I need to find some more paper though as you can only fit so much on an enevelope

[Old_Geezer]

I take it to mean that B/C = A/(A-1)

Having taken a quick look at it I don't think that the solution is as easy as I anticipated. Unsure I have the patience to work it out, but got a gut feeling manipulation should lead you to a quadratic at some point. One hopes then that the result fits the Pythagoras Theorem.

[Prudie]

I can see both too now, it depends on where you stress the word rather than the maths logic. I'm the same, at work scribbling on a jotter pad when no-one's looking so not getting very far.

[Old_Geezer]

No, I see I should think before posting, that isn't right.

B/C = 1+ (A/(A-1)

I always seem to post too early these days. Maybe I panic more in my old age.

[Old_Geezer]

Going to scream in a minute

B/C = 1+ (A/(A-1))

Yet again why is there no edit facility here. No one has come up with a legitimate excuse.

[Old_Geezer]

Yes I can now see the second interpretation but the punctuation doesn't imply that is correct.

[fiction-factory]

Both interpretations seem to lead me to very small values for C- the former gave me C

[fiction-factory]

Ah- the old problem of less than and more than signs means all text thereafter doesn't appear.

[fiction-factory]

...former gave C less than 1, latter gave me C less than 0.5

I can't face retyping everything else after that that disappeared

[fiction-factory]

I'll take a break as maths symbols are so difficult to type on here. The value of C SQUARED is always less than 0.5 so the value of C is always close to but less than half of square root of 2 .
If the sun shines I'll sit out later and have a go.

[Old_Geezer]

Shows how daft I'm getting. It has only just occurred to me that the Pythagoras Theorem is a further equation to consider at the start when solving. Not just something you check your solution against later.
Maybe I'll go back to bed.

[jim360]

There has to be a mistake in this puzzle as stated. There are three equations with three unknowns, so that we can obtain an exact solution, which is either a = 2.068, b=1.958, c=0.667 or a= 2.058, b=1.945, c=0.673 (along with several unphysical solutions in each case). Neither of these has integer value for the hypotenuse. It's not obvious to me that *any* reasonable variation of the problem will provide integer values for a.

[British_Olympics]

Thanks everyone, I did wonder whether there might be a mistake somewhere, and though not explicitly stated, I'm confident that A has to be an integer because we are later told that A squared minus 2, all squared is an integer. Shall watch out for the published result and let you know.

[British_Olympics]

Doh, as soon as I posted I realised that it's A squared rather than A that has to be an integer, but neither of jim's solutions seem to fulfil that criterion.

[fiction-factory]

If it doesn't actually say A is an integer, only that A squared -2) all squared is an integer, then it may be that A is a multiple of a square root of 2

[jim360]

Ah.

No, it doesn't have to be an integer, then. (A^2-2)^2 = a^4 - 4 a^2 + 4, which will give an integer if either a^2 is an integer, or a^2 ( a^2 -4) is integer.

It turns out that for my second solution, (a^2-2)^2 = 5 (up to rounding error), from which a = Sqrt[Sqrt[5] +2]. By magic, inserting this into the condition that a^2 (a^2 - 4 ) is integer works, as a^2 -4 is sqrt[5]-2 and a^2 is sqrt[5] + 2, so that a^2(a^2 - 4) = 5 - 4 = 1.

[jim360]

(Just to clarify, this answers I gave were rounded to 3 dp. Using the 6dp answers I got from Wolfram gave that the second one goes to 4.99998, which is as near as dammit to 5.)