Perfect Squares and Factorials

Number Theory Level 4

How many ordered pairs of positive integers ( x , y ) (x,y) satisfy x 2 + 10 ! = y 2 ? x^2 + 10! = y^2?


The answer is 105.

View wiki
Eli Ross by Eli Ross

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Eli Ross Staff
Sep 12, 2015

50 Helpful 0 Interesting 0 Brilliant 0 Confused

I also did this in the same way

Manit Kapoor - 5 years, 9 months ago

(8+1)....wat(6+1)?

Soner Karaca - 5 years, 7 months ago

Log in to reply

Because a and b should be even. So, e.g. a can have 2^1 up to 2^7 ... thus (6+1) choice.

Pierre Carrette - 3 years ago
Vinod Kumar
Oct 15, 2018

Used Wolfram Alpha to solve Diophantine equation x^2-y^2=10! to get (420/4)=105 positive integer solutions.

Answer=105

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...