## Search found 55 matches

Mon Apr 25, 2016 11:31 pm
Forum: Number Theory
Topic: Integer solution
Replies: 5
Views: 596

### Re: Integer solution

SOLUTION 1: From the equation, we can easily see, x & y are of different parity. So, Case 1 : At first,we let, x = 2p and y=2q+1 then, from equation, \$4q^2\$+4q +1 = \$8p^3\$ + 7 => \$4(q^2 +q)\$ = 4(\$ 2p^3\$ +1) +2 .Here, 4 divides LHS but not RHS.Contradiction! Case 2: Here, x=2p+1 ,y=2q as 2nd case. W...
Sat Apr 23, 2016 4:16 pm
Forum: Number Theory
Topic: International Zhautykov Olympiad 2005, Problem 6
Replies: 2
Views: 592

### Re: International Zhautykov Olympiad 2005, Problem 6

\$pq\$ divides \$p^2q^2 +8(p^2 +q^2)+64\$ so, \$pq\$ divides \$8(p^2 +q^2 +8)\$ so, \$p\$ divides \$q^2\$ and \$q\$ divides \$p^2\$ Ahmed Ittihad ,Your approach is wrong.your start was statisfactory indeed; but in your 3rd line, u made a mistake. look, u told in ur 3rd line that (\$p\$ divides \$q^2\$ and \$q\$ divides ...
Wed Apr 20, 2016 5:49 pm
Forum: Combinatorics
Topic: Binomials
Replies: 3
Views: 751

### Re: Binomials

[I assume,u understand case1,as u wanted case2's explaination.].. Case2 mainly tells that,u've to take xyz such that the product is not a multiple of 9, but a multiple of 3(as case2,we find out when the product is neither multiple of 3 nor 9 in case1).So, we r now finding out the quantity of product...
Wed Apr 06, 2016 4:12 pm
Forum: Combinatorics
Replies: 2
Views: 437