A&B

Algebra Level pending

a 3 + ( a + 1 ) 3 + ( a + 2 ) 3 + ( a + 3 ) 3 + ( a + 4 ) 3 + ( a + 5 ) 3 + ( a + 6 ) 3 = b 4 + ( b + 1 ) 4 a^{3} + (a+1)^{3} +(a+2)^{3} +(a+3)^{3}+(a+4)^{3}+(a+5)^{3}+(a+6)^{3}= b^{4} +(b+1)^{4}

What is the number of solutions ( a , b ) (a,b) satisfying the equation above?


The answer is 0.

Priyansh Singh by PRIYANSH SINGH , 19, India

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.

1 solution

Priyansh Singh
Oct 2, 2018

In the case of the LHS we can try modulo 7 and can write it that it is divisible by 7.since LHS can be expressed as 0(mod 7)

Now for RHS it can never be expressed as 0(mod 7) and hence no solutions exist

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...