m^6+5n^2=m+n^3

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m^6+5n^2=m+n^3

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Source: Turkey JBMO TST 2014 P3

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#1 h Jun 21, 2014, 2:02 PM • 2 Y

Y by Adventure10, Mango247

Find all pairs $(m, n)$ of positive integers satsifying $m^6+5n^2=m+n^3$ .

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#2 h Jun 22, 2014, 3:00 AM • 2 Y

Y by Adventure10, Mango247

This is the solution for all integer numbers $m,n \ge 0$ .
The equation is equivalent to $(m^2-n)(m^4+m^2n+n^2)=m-5n^2.$
Case 1. If $m=5n^2$ then $m^4+m^2n+n^2=0$ or $m^2-n=0$ .
If $m^4+m^2n+n^2=0$ then $m=n=0$ .
If $m^2=n$ , we also have $m=5n^2$ then $25n^4=n$ . We get $n=m=0$ .
Case 2. If $m>5n^2$ . It follows that $m-5n^2 \ge m^4+m^2n+n^2$ or $m \ge m^4+m^2n+6n^2$ . Since $m^4 \ge m$ so $m=n=0$ , a contradiction.
Case 3. If $m<5n^2$ . With $m=0$ then $n^2(n-5)=0$ . We obtain $n=5$ .
With $m \ge 1$ , from the equation we get $\frac{5n^2-m}{m^4+m^2n+n^2}=n-m^2=k \ge 1$ . The equation become $\begin{array}{l} (5-k)n^2-nm^2k-m^4k-m=0 \qquad (1) \\ \Delta= (m^2k)^2+4(5-k)(m^4k+m)=m^4(20k-3k^2)+4m(5-k). \end{array}$
Equation $(1)$ has roots when $\Delta \ge 0$ . It is easy to see that if $k \ge 7$ then $\Delta < 0$ . Therefore $1 \le k \le 6$ .

If $k=6$ then $n=m^2+6$ . Hence $6[m^4+m^2(m^2+6)+(m^2+6)^2]=5(m^2+6)^2-m$ or $m(13m^3+48m+1)=-36$ , a contradiction since $LHS >0> RHS$ .

If $k=5$ then $n=m^2+5$ . Hence $5[m^4+m^2(m^2+5)+(m^2+5)^2]=5(m^2+5)^2-m$ or $m(10m^3+25m+1)=0$ , a contradiction since $LHS>0$ .

If $k=4$ then $n=m^2+4$ . Hence $4[m^4+m^2(m^2+4)+(m^2+4)^2]=5(m^2+4)^2-m$ or $m(7m^3+8m+1)=2^4$ . Since $\gcd (m,7m^3+8m+1)=1$ and $m<7m^3+8m+1$ so $2 \nmid m$ . We find $m=1$ . Hence $n=5$ .

If $k=3$ then $n=m^2+3$ . It follows that $3[m^4+m^2(m^2+3)+(m^2+3)^2]=5(m^2+3)^2-m$ or $m(4m^3-3m+1)=18$ . We see that $m \ne 1$ . If $m \ge 2$ then $LHS>18$ , a contradiction.

If $k=2$ then $n=m^2+2$ . It follows that $2[m^4+m^2(m^2+2)+(m^2+2)^2]=5(m^2+2)^2-m$ or $m(m^3-8m+1)=12$ . Since if $m \ge 4$ then $LHS>RHS$ then $m \le 3$ . We find $m=3$ . Ir follows that $n=11$ .

If $k=1$ then $n=m^2+1$ . Hence $m^4+m^2(m^2+1)+(m^2+1)^2=5(m^2+1)^2-m$ or $m(4m^3-7m+1)=4$ . If $m \ge 2$ then $LHS>RHS$ and $m \ne 1$ .

Thus, the answer is $(m,n)=(0;0),(0;5),(1;5),(3;11)$ . $\blacksquare$

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#3 h Jun 23, 2014, 3:35 PM • 3 Y

Y by TheRealRuirui, Adventure10, Mango247

Notice that for $n>11$ we have (with using $m<n$ )
$(3n-5)^3>(3m^2)^3=27(n^3-5n^2+m)>(3n-6)^3$
So no solution for $n>11$ .

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#4 h Jan 11, 2017, 3:07 PM • 3 Y

Y by Adventure10, Mango247, MuradSafarli

Another simple solution: from the condition, number m^6-m is of the form n^3-5n^2. For m>=4 we have that (m^2+1)^3-5(m^2+1)<m^6-m<(m^2+2)^3-5(m^2+2), so no solutions because n^3-5n^2 is strictly increasing in naturals which are >=5. After checking other cases we get that (m,n)={(1,5),(3,11)}.

This post has been edited 1 time. Last edited by MilosMilicev, Jan 13, 2017, 6:01 PM

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#5 h Aug 8, 2022, 3:19 AM • 1 Y

Y by MuradSafarli

shinichiman wrote:

Thus, the answer is $(m,n)=(0;0),(0;5),(1;5),(3;11)$ . $\blacksquare$

but $m, n \ne 0$

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#6 h Apr 9, 2025, 2:39 PM

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We are going to solve this by bounding
Motivation for the bounding
The 2 very powerful lemmas
Finishing the problem
A very great problemmm :-D

:-D

This post has been edited 1 time. Last edited by TopGbulliedU, Apr 9, 2025, 2:50 PM

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