POTW 4!

POTW 4: n=95.

I reasoned as follows: (n-12)/(5n+23) = rp/sp, where n, p, r, and s are integers, and p>1 (to make the ratio reducible). Therefore,
n=rp+12 and 5n+23=sp. Substituting the former into the latter gives 5rp+83=sp, so p must divide 83. The minimum allowable value for p is therefore 83. Then 5r+1 = s, so the minimum values for r and s are 1 and 6. Then n=(1)(83)+12 =95, and

(n-12)/(5n+23) = 83/498 = r/s = 1/6

Since 83 is prime, are there any other solutions to this besides n=95 and n=-71 (supposing n were allowed to be negative)?

@Another Matt:
Every n = 83k+12 gives a valid solution, since the fraction is then reducible by 83.

Oh, duh. That's what I get for doing these on no sleep!

I used congruences and modular arithmetic to arrive at the same answer. From the problem, it is clear that we are seeking a value for n such that n-12 = 5n+23 = 0 mod q for some modulus q (read the = signs as "tri-bar" congruence signs here).

These congruences imply that n=12 mod q and 5n = -23 mod q. the second implies 5n = q-23 mod q. Setting n =12 gives 5n = 60. Since 5n=q-23, that implies q=83. So our congruences are then n=12 mod 83 and 5n = 60 mod 83. We can't just take n=12, however, because that gives zero for the fraction. Thus, we have to take the next highest value of n that satisfies our congruences, namely n = 12+83 = 95. That then gives (95-12)/(5*95+23) = 83/498 = 1/6.

BTW, Another Matt, it is clear from my solution that any number of the form 83c+12, for n>=1 is a valid solution to the problem, for 83c+12 = 12 mod 83 and 5*(83c+12) = 60 mod 83. (Both of these follow easily from the fact that 83c = 0 mod 83).

It is similarly clear from my work above that n=83r+12 is a solution for any r, with the minimum at r=1. Had this been part of the problem assignment I would have noted it.

POTW 5: I think n=6, (n-6)(n+14)=0=0^2 is the only solution for positive n (n=-14 is the only solution for negative n).

(n-6)(n+14)=m^2 implies that any non-zero m has two factors (which could be itself and 1) = a^2(b^2) where a and b are two non-zero integers and I will take a to be the smaller one (a<b).

Case 1: n-6 = a^2 and n+14 = b^2. Then b^2 - a^2 = 20 and so a and b must have the same parity (both be odd or both be even). This factors into (b+a)(b-a)=20, so (b+a) and (b-a) must be factors of 20, either 20 and 1 or 5 and 4; but since b and a must be the same parity, both factors must also be the same parity, which is not true in either case, so no solution of this form is possible.

Case 2: n-6=a and n+14=a(b^2)). Then n=a+6. and substituting this into the second equation gives a+20 = a(b^2), so a must be a factor of 20 - either 1, 20, 4, or 5. None of these gives a solution in which b^2 is a square integer, so there are no solutions for this case either.

Case 3: n-6=b and n+14=(a^2)b - then b+20 = (a^2)b, which parses the same way as for Case 2.

Since a<b and n-6 < n+14, there is no other way to assign the factors, and m^2 =0 is the only possible solution.

POTW 5 is nice.

We have the hyperbola y^2 = x^2 + 8x - 84. Complete the square and move a few things around:

y^2 = (x^2 + 8x + 16) - 100

y^2 = (x+4)^2 - 100

(x+4)^2 - y^2 = 100

for simplicity substitute n = x + 4)

n^2 - y^2 = 100

(n-y)(n+y) = 100

So we need to find an n and a y such that n-y and n+y are mutual factors of 100. 100 factors into primes 2*2*5*5, with pairs 2*50, 4*25, and 10*10.

For the 4*25 pair, we have n=14.5 and y = 10.5, but these are not integers, so there is no solution.

For the 10*10 pair, we have n=10 and y=0, so x=n-4=6.

For the 2*50 pair, we have n=26 and y=24, so x=n-4=22.

Plugging them into the original formula to verify we get:
(6-6)(6+14) = 0 = 0^2

and

(22-6)*(22+14) = 576 = 24^2

JimV, looking at your case 1, you forgot the 10*2 factor, which does have the same parity. In this case a^2=16 and b^2=36, and the resulting n = 22.

x=-30 is the other negative answer.

Very good work, A. Matt. I missed the 20=2(10) factors, as you said.

This time I got some sleep! I have twin babies at home, so doing anything mathy can be precarious.

In my example I also missed the 20*5=100 factor pair, but since they aren't the same parity, there would be no solution.

Kudos, AM on an elegant solution. I struggled with this problem and didn't find a solution until I gave up and read yours. For completeness, though, you should have also considered the pair 1 and 100, which of course leads to a non-integer value, so your solution still stands as correct.

Yep, thanks. It's better formed using the obvious parity constraint, which means that the two 2's in the prime factors have to belong to different factors in the pairs. Then all you have to consider is 10*10 and 2*50. All of the others (1*100, 4*25, 5*20) necessarily have opposite parity, and can be rejected without enumeration.