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no numbers of the form 80...01 are squares
Marius_Avion_De_Vanatoare   2
N Apr 3, 2025 by EVKV
Source: Moldova JTST 2024 P5
Prove that a number of the form $80\dots01$ (there is at least 1 zero) can't be a perfect square.
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Marius_Avion_De_Vanatoare
Jun 10, 2024
EVKV
Apr 3, 2025
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no numbers of the form 80...01 are squares
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Source: Moldova JTST 2024 P5
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Marius_Avion_De_Vanatoare
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Marius_Avion_De_Vanatoare
#1 h  Jun 10, 2024, 2:37 PM
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Prove that a number of the form $80\dots01$ (there is at least 1 zero) can't be a perfect square.
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navier3072
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navier3072
#2 h  Jun 10, 2024, 3:00 PM
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Suppose not. Then, for $k\geq 2$, $80\dots01=8 \cdot 10^k +1=X^2$. Then, \[2^{k+3} \cdot 5^k =(X+1)(X-1) \]Hence,
Case 1: $2^{k+2} \cdot 5^m - 2 \cdot 5^n =2 $
Then, $2^{k+1} 5^m - 5^n =1 $. Since, $n>m$ (by size arguments), $5^m \equiv 1 \pmod 5 \implies m=0$, $n=k$. Thus, $2^{k+1} - 5^{k} =1 $, contradiction by, say, Mihailescu.

Case 2: $2 \cdot 5^m - 2^{k+2} \cdot 5^n =2 $
Then, using similar arguments, since $m>n$, $n=0$ and $m=k$, so $5^{k} - 2^{k+1} =1 $, contradiction.
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EVKV
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EVKV
#3 h  Apr 3, 2025, 2:21 AM
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FTSOC assume it is possible for some k
$2^{k+3}5^{k}$ +$1$ = $X^{2}$
As $gcd(X+1,X-1) = 2$ Only 2 cases possible
Case 1
$2^{k+2}5^{k} -2 = 2$
which is nonsense modulo 5
Case 2
$25^{k} - 2^{k+2}= 2$
again nonsense for $k \geq 2$ (By Catalans or if u want induction)

For $k = 1 $
801 is not a square
This post has been edited 1 time. Last edited by EVKV, Apr 3, 2025, 2:23 AM
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