see https://www.artofproblemsolving.com/community/c6h1245722_primes_number

JBMO SL 2015 N5 wrote: Check if there exists positive integers $ a, b$ and prime number $p$ such that $a^3-b^3=4p^2$ Observe that the positive divisors of RHS are $1,2,4,p,2p,4p,p^2,2p^2,4p^2$ and hence we'll consider cases for values of $a-b$ and $a^2+ab+b^2$ $\bullet$ If $a-b=p^2$, then, $a^2+ab+b^2=4$, which has no positive integer solutions for $a,b$ $\bullet$ If $a-b=2p^2$, then $a^2+ab+b^2=2$, again no solution $\bullet$ If $a-b=4p^2$, then $a^2+ab+b^2=1$ again has no positive integer solution! $\bullet$ If $a-b=1$ $\implies$ $a^2+ab+b^2=4p^2$ $$a=b+1 \implies a^2+ab+b^2=(a-b)^2+3ab=1+3ab=4p^2 \implies 3b^2+3b+1=4p^2$$$$3b^2+3b+1 \equiv 0 \pmod 4 \implies b(b+1) \equiv 1 \pmod 4$$which isn't possible! $\bullet$ If $a-b=2$ $\implies$ $a^2+ab+b^2=2p^2$ $$a=b+2 \implies a^2+ab+b^2=(a-b)^2+3ab=4+3b^2+6b=2p^2 \implies 2 | 3b^2 \implies b=2k $$$$12k^2+4k+4=2p^2 \implies p=2 \implies \text{ No integer solutions}$$$\bullet$ If $a-b=4$ $\implies$ $a^2+ab+b^2=p^2$ $$a=b+4 \implies a^2+ab+b^2=(a-b)^2+3ab=16+3b^2+12b=p^2$$If $p=2$, we have no solution! If $p>3$,then, $$p^2 \equiv 1 \equiv 3b^2+12b+16 \equiv 3b^2 \equiv -b^2 \pmod 4 \implies b^2\equiv 3 \pmod 4$$No solution! $\bullet$ If $a-b=p$ $\implies$ $a^2+ab+b^2=4p$ $$a^2+ab+b^2=4(a-b) \implies (a-b)(a-b-4)=p(p-4)=-3ab \implies p=2,3 $$But $p=a-b=2$ is not possible hence, if $p=3$ $$a-b=3 \implies a^2+ab+b^2=(a-b)^2+3ab=9+3ab=12 \implies a=b=1 \text{ which is not a solution}$$$\bullet$ If $a-b=2p$ $$\implies a^2+ab+b^2=(a-b)^2+3ab=a-b$$$$(a-b)(a-b-1)=2p(2p-1)=-3ab$$Hence we have no solution! $\bullet$ If $a-b=4p$ $\implies$ $a^2+ab+b^2=p$ $$\implies a-b=4(a-b)^2+12ab \implies -12ab=(a-b)(4(a-b)-1)=4p(16p-1) $$Again No solution!

An easier solution in my opinion is Taking mod 4we find that a=b mod4 If not then a^2+b^2+ab is 0 mod 4 a^2+b^2 is 1mod 4 Since a,b are different mod 4 and ab is 0,2 mod 4 so impossible then let (a-b)/4=x Then $x(a^2+b^2+ab)=p^2$ taking 2 cases we find no sol.

What about $4|a-b-2,\ 4|a^2+b^2+ab-2$? You didn't consider this case.

WolfusA wrote: What about $4|a-b-2,\ 4|a^2+b^2+ab-2$? You didn't consider this case.

I think it is much easier: By taking mod 4 we observe that LHS is 0 mod4 thus a,b must be both even (if both odd... no solution). So a=2k and b=2n. Now, we easily get a^3-b^3=0 mod 8 => p=2!!! The equation becomes (a-b)(a^2+ab+b^2)=16. We consider two cases: 1) (a-b, a^2+ab+b^2) = (1, 16) ......no solution 2) (a-b, a^2+ab+b^2) = (2, 8) .......no solution

sttsmet wrote: I think it is much easier: By taking mod 4 we observe that LHS is 0 mod4 thus a,b must be both even (if both odd... no solution). So a=2k and b=2n. Now, we easily get a^3-b^3=0 mod 8 => p=2!!! The equation becomes (a-b)(a^2+ab+b^2)=16. We consider two cases: 1) (a-b, a^2+ab+b^2) = (1, 16) ......no solution 2) (a-b, a^2+ab+b^2) = (2, 8) .......no solution nice idea!

if $a,b =0 \pmod{2}$ $\Rightarrow$ $a^3-b^3 =0 \pmod {2}$ but LS dont give so $a,b =1\pmod {2}$ $\dfrac{a+b}{4} (a^2+ab+b^2)=p^2$ $a^2+b^2+ab=p^2,p$,$\dfrac{a+b}{4}=p,1$ $\dfrac{a+b}{4}=1$$(a,b)=(3,1),(2,2),(1,3)$ with this $a^2+ab+b^2\ne p^2$ $a^2+b^2+ab=p$,$\dfrac{a+b}{4}=p$ $a+b=4(a^2+ab+b^2)$but its impossible

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