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Source: Pan African Mathematics Olympiad p6

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fasttrust_12-mn

118 posts

fasttrust_12-mn

#1 h Aug 16, 2024, 10:21 AM • 1 Y

Y by cheikhennahoui

Find all integers $n$ for which $n^7-41$ is the square of an integer

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678 posts

#2 h Aug 16, 2024, 10:43 AM • 2 Y

Y by cheikhennahoui, Monica_Twigg_239

We can rewrite it as $n^7+2^7=m^2+13^2$ and use the approach from USA TST 2008 P4.

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DylanN

#4 h Aug 16, 2024, 2:42 PM • 1 Y

Y by Monica_Twigg_239

I personally probably would have waited until the paper was finished being written before posting the problems online.

To fill in the details for the other response, we note that $m^2 + 13^2$ is a sum of two squares, and so the only possible primes factors are $2$ , prime factors of $\gcd(m, 13)$ , and primes that are congruent to $1$ modulo $4$ . If $2$ were one of the prime factors, then $n$ would be even. But then we would have that $n^7 - 41 \equiv 3 \pmod 4$ , and so it would not be a square. Thus every prime factor of $n^7 + 2^7$ is congruent to $1$ modulo $4$ . (Since $13$ is itself congruent to $1$ modulo $4$ .) Since $n + 2$ is a factor of $n^7 + 2^7$ , this implies that every prime factor of $n + 2$ is congruent to $1$ modulo $4$ , and so $n + 2 \equiv 1 \pmod 4 \implies n \equiv -1 \pmod 4$ . But then $n^7 + 2^7 \equiv (-1)^7 \equiv 3 \pmod 4$ , and so $n^7 + 2^7$ would have a prime factor congruent to $3$ modulo $4$ , a contradiction.

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28 posts

#5 h Aug 16, 2024, 5:32 PM • 1 Y

Y by Monica_Twigg_239

yeah same old trick as USATST 2008 p4

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#6 h Apr 23, 2025, 10:31 PM • 1 Y

Y by ATM_

Motivation (USATST p4 trick: $n^7-7=m^2$

Notice that : $41=13^2-2^7$
So : $n^7+2^7=13^2$
And : $n+2|n^7+2^7\implies n+2|m^2+13^2$
By Fermat's Chrismats theoreme : $n+2\equiv 1[4]\implies n\equiv -1[4]\implies n^7+2^7\equiv -1[4]\implies m^2+13^2\equiv -1[4]$
Hence : $m^2\equiv 2[4]$
Which is impossible because quadratique residus mod 4 are 0,1

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270 posts

#7 h Apr 23, 2025, 10:33 PM

Y by

known trick, like BMO 1998

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