Difference between revisions of "2007 JBMO Problems/Problem 4"
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==Problem 4== | ==Problem 4== | ||
Prove that if <math>p</math> is a prime number, then <math>7p+3^{p}-4</math> is not a perfect square. | Prove that if <math>p</math> is a prime number, then <math>7p+3^{p}-4</math> is not a perfect square. | ||
| + | <math>\int</math> | ||
| + | |||
==Solution== | ==Solution== | ||
By Fermat's Little Theorem, <math>7p+3^p-4\equiv3-4\equiv-1\mod p</math>. By quadratic residues, this is true if and only if <math>p\equiv1\mod4</math>, except for <math>p=2</math> (which doesn't work). Then, <math>7p+3^p-4\equiv3+3-4=2\mod4</math>, but this implies <math>v_2(7p+3^{p}-4)</math> is odd, so <math>7p+3^{p}-4</math> cannot be a perfect square. | By Fermat's Little Theorem, <math>7p+3^p-4\equiv3-4\equiv-1\mod p</math>. By quadratic residues, this is true if and only if <math>p\equiv1\mod4</math>, except for <math>p=2</math> (which doesn't work). Then, <math>7p+3^p-4\equiv3+3-4=2\mod4</math>, but this implies <math>v_2(7p+3^{p}-4)</math> is odd, so <math>7p+3^{p}-4</math> cannot be a perfect square. | ||
Revision as of 14:30, 16 April 2024
Problem 4
Prove that if 
is a prime number, then 
is not a perfect square.
Solution
By Fermat's Little Theorem, 
. By quadratic residues, this is true if and only if 
, except for 
(which doesn't work). Then, 
, but this implies 
is odd, so cannot be a perfect square.
See also
| 2007 JBMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
