The primary drawback on the eighty fifth Putnam was a enjoyable and never too tough query. (I nonetheless didn’t resolve it, however the answer is comparatively easy.)
Decide all constructive integers n for which there exist constructive integers a, b, and c satisfying:
2an + 3bn = 4cn
As traditional, watch the video for an answer.
Or preserve studying.
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“All will probably be nicely when you use your thoughts in your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of recreation concept and arithmetic. MindYourDecisions now has over 1,000 free articles with no advertisements because of group help! Assist out and get early entry to posts with a pledge on Patreon.
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Reply To Simple Query From 2024 Putnam
(Just about all posts are transcribed shortly after I make the movies for them–please let me know if there are any typos/errors and I’ll appropriate them, thanks).
Due to Tony for alerting me of a typo!
The reply is the equation has an answer in constructive integers for n = 1 solely.
Beginning
Begin with n = 1. With some experimenting, there’s a answer (a, b, c) = (1, 2, 2).
2(1) + 3(2) = 4(2) = 8
In my thoughts, the subsequent step was to consider Fermat’s well-known conjecture, confirmed by Wiles in 1995, that there isn’t any answer in constructive integers for n ≥ 3 for the equation:
xn + yn = zn
So I’d assume to think about n ≥ 3 and present there isn’t any answer.
Comparatively prime
Suppose as a substitute (a, b, c) is an answer. Let d = gcd(a, b, c). Outline x = a/d, y = b/d, and z = c/d. Then we’ve got (x, y, z) is an answer as a result of:
2xn + 3yn
= 2(a/d)n + 3(b/d)n
= (1/dn)(2an + 3b)n
Since (a, b, c) is an answer, we’ve got:
= (1/dn)4cn
= 4(c/d)n
= 4zn
Thus (x, y, z) is an answer, and they’re comparatively prime.
We will subsequently assume with out lack of generality we’ve got an answer the place the numbers are comparatively prime and gcd(a, b, c) = 1.
Case n ≥ 3
Suppose n ≥ 3.
2an + 3bn = 4cn
3bn = 4cn – 2an
Since 2 divides the precise hand aspect of the equation, 2 divides 3bn. However 2 doesn’t divide 3, and since 2 is prime, we will conclude 2 divides b, so b = 2okay. Substituting to the unique equation provides:
2an + 3(2okayn) = 4cn
an = 2cn – 2n – 1 (3okayn)
Since n ≥ 3, the precise hand aspect is divisible by 2, which implies an is divisible by 2, and a = 2m. Substituting we get:
(2m)n = 2cn – 2n – 1 (3okayn)
2n – 1mn + 2n – 2(3okayn) = cn
Since n ≥ 3, the left hand aspect is divisible by 2, which implies cn is divisible by 2, and c = 2p.
However we now have proven a, b, c are all divisible by 2, which contradicts gcd(a, b, c) = 1.
We sadly can not simply use this identical proof for n = 2, as a result of then we’ve got the time period:
2n – 2(3okayn)
= 20(3okayn)
= 3okayn
We can not conclude that is divisible by 2. So we’ve got to deal with the case of n = 2 individually.
Case n = 2
Suppose as earlier than we’ve got an answer (a, b, c) and gcd(a, b, c) = 1. For n = 2 the equation turns into:
2a2 + 3b2 = 4c2
Working modulo 3 we get:
–a2 ≡ c2 (mod 3)
0 ≡ a2 + c2 (mod 3)
a2 + c2 ≡ 0 (mod 3)
The one squares modulo 3 are 0 and 1 as a result of:
02 ≡ 0 (mod 3)
12 ≡ 1 (mod 3)
22 ≡ 1 (mod 3)
So the sum of two squares could possibly be solely 4 prospects:
0 + 0 ≡ 0 (mod 3)
1 + 0 ≡ 1 (mod 3)
0 + 1 ≡ 1 (mod 3)
1 + 1 ≡ 2 (mod 3)
Solely the primary case corresponds to a sum of 0 (mod 3), so we will need to have each squares are identically equal to 0 (mod 3).
a2 ≡ c2 ≡ 0 (mod 3)
Since 3 is prime, this suggests a and c are each divisible by 3. That’s, a = 3v and c = 3w.
Substituting to the unique equation we’ve got:
2(3v)2 + 3b2 = 4(3w2)
3b2 = 4(3w2) – 2(3v)2
3b2 = 36w2 – 18v2
b2 = 12w2 – 6v2
b2 = 3(4)w2 – 3(2)v2
Since 3 divides the precise hand aspect, it should divide b2, so 3 divides b.
However we’ve got now proven 3 divides every of a, b and c, which contradicts gcd(a, b, c) = 1.
The reply is the equation has an answer in constructive integers for n = 1 solely.
References
official options
https://maa.org/wp-content/uploads/2025/02/2024-Putnam-Session-A-Options-1.pdf
unofficial options
https://kskedlaya.org/putnam-archive/2024s.pdf
stamp
https://en.wikipedia.org/wiki/Fermatpercent27s_Last_Theorem#/media/File:Czech_stamp_2000_m259.jpg
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