by Bet667, Apr 3, 2025, 6:54 AM 
be real numbers less than 2.Then prove that 
by SunnyEvan, Apr 1, 2025, 9:54 AM 
, 
, 
be non-negative real numbers, no two of which are zero. Prove that :
by steven_zhang123, Mar 29, 2025, 12:42 AM 
be an odd integer, 
. Consider 
integers 
such that for any 
, 
and 
. Prove that the number of 
for which satisfy 
for some 
, 
is greater than 
.by Rainbow1971, Mar 28, 2025, 8:39 PM 
be an integer. Show that, if 
is an integer, then it is a perfect square.
. If is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?by KAME06, Nov 4, 2024, 8:26 PM 
be a 90-degree triangle with hypotenuse 
. Let 
and 
distinct points on segment and 
be the foot of the perpendicular from to 
and to 
, respectively. 
and 
intersect at 
.
and intersect at 
and lines 
and intersect at 
.
if and only if 
.by David-Vieta, Apr 1, 2023, 10:31 AM be a positive integer. Initially, a 
grid has 
black cells and the rest white cells. The following two operations are allowed :
square has exactly three black cells, the fourth is changed to a black cell;
square, the black cells are changed to white and white to black. such that for any configuration of the grid with black cells, all cells can be black after a finite number of operations.by xeroxia, Sep 27, 2006, 10:06 AM 
of positive integers such that ![\[t^{2}+1=s(s+1).\]](http://latex.artofproblemsolving.com/f/6/6/f665bfe4fd6cec17dbc5b90d42d201274ff7c564.png)
The oldest, shortest words — "yes" and "no" — are those which require the most thought.
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and 
: Prove that :
and 
Prove that
April 2021